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【典中点】2017春浙教版八年级数学下册1. Integrated promotion password 2'/>Downloads [典中点]2017春浙教版八年级数学下册1. Integrated upgrade password decoding Special training one: use the concept of the quadratic root formula to find letters or algebra, and use the definition of the quadratic root formula to determine the quadratic root formula 1. The following formula is not necessarily the quadratic root type ( ) A. (x ≤ 0) D Use the second root meaningful condition to find the value range of the letter 2. Regardless of the real number of x, the algebraic expression is meaningful, and the simplified form +. uses the definition of the simplest quadratic formula to identify the simplest quadratic root 3. Which of the following quadratic roots are the simplest quadratic roots? Which ones are not? Why?,,, (x>2), -x,, (b>0, a>0),, (a>b>0) ,,.4. Simplify the following formulas: (1); (2) (a ≥ 0, b ≥ 0); (3) (mn > 0); (4) (x ≠ y). For the same simplest quadratic condition, find the value of the letter 5. If the simplest root is the simplest quadratic root of the same number of squares, then ( ) A.a=0, b=2B.a=2 , b=0C.a=-1, b=1D.a=1, b=-26 If the simplest quadratic roots can be combined, the algebraic form -+(3a+2b)2 has the value ________.7. If the simplest quadratic root can be combined with the quadratic root addition and subtraction, find a meaningful x The value range is 8.8. If m, n are rational numbers, and ++=m+n, find the value of (m-n)2+2n. Decoding 2: Compare the eight methods of the secondary root size,,,, and the flat method 1. Compare the size of + and +. As a business method 2. Compare the size of 4- and 2+. Molecular physicochemical method 3. Compare - and - size. Denominator physicochemical method 4. Compare the size of the difference. Compare the size of the sum. Reciprocal method 6. Know x = -, y = -, try to compare the size of x, y. Special value method 7. Use '0, y> 0, z > 0), find the value. The first algorithm is 11. It is known that a+b=-8, ab=8, simplifies b+a and evaluates. Decoding Specialized 4: Using the double non-negative simplification of the quadratic root, 'non-negative': a ≥ 0, ≥ 0, 'non-negative' is often used in solving the problem of the quadratic root. The number of squares and values ??of the quadratic root are non-negative, and the problem of the quadratic root is solved by the number of squares a≥0. If -=, the value of 3x-y is ________. Use ≥0 to find the value of algebra. 2|2a-4|+=0 is known, and the value of a+b-ab is obtained. Use ≥ 0 to find the maximum value 3. When x is taken, what is the value of +3? What is the minimum value? Using the non-negative solution of the number of squares to solve the algebraic simplification problem. Let the equation +=- be established, and x, y, a are not equal to each other, and find the value. Decoding Specialized Five: Using the quadratic root solution to solve the problem related to a right triangle, use the quadratic root to find the length of the line segment in the right triangle. As shown in the figure, △ABC, ∠ACB=90°, AB=, BC=, find the high CD on the oblique side AB. (Question 1) Use the quadratic root formula to find the perimeter and area of ??the quadrilateral. The upper base of a right-angled trapezoid is 2cm, the bottom is cm, and the height is cm. Find the area and circumference of the trapezoid. Using the quadratic root formula to solve the coordinates of the point in the plane rectangular coordinate system. It is known that in RtΔOAB, ∠B=90°, the coordinate of point A is (,0), BA=2. Place △OAB in the Cartesian coordinate system as shown in the figure, so that point O coincides with the origin, point A falls on the positive half of the x-axis. Find the coordinates of point B. (Question 3) Solving the actual problem using the quadratic root formula 4. As shown in the figure, there is a pumping station A 10m in the northeast direction of the water tower O, a construction site B 20m in the southeast direction of the water tower, a straight water pipe is laid between the AB, and the length of the water pipe is sought. (Question 4) Decoding training Six: Five popular test sites in the secondary root formula, and the second root type of meaningful conditions and properties 1. (中考·南京) If the expression is meaningful within the real range, the value range of x is ________. 2. (中考·黔南州) The position of the corresponding point of the real number a on the numerical axis is as shown in the figure, simplification + a = ________. (Question 2) 3. If it is opposite to each other, find the square root of 6x+y. Simplification and operation of the quadratic roots 4. (中考·徐州) The following errors in the operation are ( ) A.+= B.×=C.÷=2D. (-) 2=35. If the simplest root can be combined, then 2a+3b=________.6. (中考·张家界) Calculation: (-1)(+1)-+|1-|-(π-2)0+. Simplification of the quadratic roots. (中考·呼和浩特) Simplify first and then evaluate: +÷, where a=, b=-.8. (School entrance exam Jingmen) first simplify, and then evaluate: +÷, where a, b meet +|b-|=. The length of one side of the isosceles triangle is 2, and the circumference is 4+7. Find the waist length of this isosceles triangle. 10. As shown in the figure, the ratio of the slope to the water in the section of the reservoir dam (the ratio of the length of DE to AE) is 5:3, the ratio of the backwater slope is 1:2, the height of the dam is DE=30m, and the width of the dam is CD=10m. Find the section perimeter of the dam. (Question 10) The regularity of the quadratic roots. (中考·Heze) The following is a sequence arranged according to a certain rule: according to the law of array arrangement, the nth (n is an integer, and n≥3) rows from left to right n-2 numbers are __________. (Use algebraic expressions with n) Decoding Specialist 7: Thoughts and methods of meta-classification discussion ideas,,,,,,,1. It is known that a is a real number and a value of -. Masters point gold: the combination of numbers and shapes is the intrinsic link between the questions and conclusions based on mathematical problems,,,,,,, 2. It is known that the position of the corresponding point of the real number m, n on the number axis is as shown in the figure, simplification: +++-. (the second question) analogy thought,,,,,,, 3. Calculation: (7+2-)(2-7+). Transforming ideas,,, 4. Calculation: (+)2015·(-)2016. Answer decoding training one. D dial:,, (x ≤ 0) is a quadratic root, can be turned into, only when x = 4, is the quadratic root, it is not necessarily the secondary root. 2. Solution: ∵ =, and no matter what real number x takes, the algebraic formula is meaningful, ∴m-4≥0, ∴m≥4. When m≥4, +=(m-3)+(m-4)=2m -. Solution:,,, is the simplest quadratic. , (x>2), -x, (b>0, a>0), (a>b>0), not the simplest quadratic. ∵===9,==x-2(x>2), -x==-,==,=b(b>0,a>0),=(a+b)(a>b>0), =, 4. Solution: (1) ==. (2) == 2a (a ≥ 0, b ≥ 0). (3) From -≥0, mn>0, m<0, n<0, ∴===-(mn>0). (4) == (x≠y). 5. A dial: the title is solved by the title. 1 Dial: ∵ The simplest quadratic root can be combined, ∴5a+b=2a-b, ∴3a+2b=0, ∴3a=-2b.∴-+(3a+2b)2=1+0=. Solution: The meaning of the question is 3a-8=17-2a.∴a=5.∴=. To make sense, just make sense. ∴20-2x≥0, ∴x≤. Solution: ∵++=+2+==m+n, ∴m=0, n=.∴(m-n)2+2n=+2×=+7=. Decoding Specialist II 1. Solution: Because (+)2=17+2, (+)2=17+2, 17+2>17+2, so (+)2>(+)2, and because +>0, +>0, so +>+.2. Solution: ∵=(4-)(2-)=11-6,6≈,∴11-6<1, then ∵4->0,2+>0,∴4-<2+.3. Solution: -==, -==, ∵+>+, +>0, +>0, ∴0, so >0, so >.6. Solution: ==>0,==>0, ∵+>+>0, ∴>>0, ∴x<. Solution: Take the special value x=, then x2=,=,=4, ∴x2. Decoding Specialist III 1. C: dial: original = 4 × + 3 = 2+3 = 5. ∵≈, ∴ 5 ≈∵ 7 < < 8, select the point: because - <0, 2 << 3, 3 << 4, so covered by ink The number is .3. Solution: original = (5 +) × [5 - () 2 ×] = (5 +) × [ × (5 -)] = × (5 +) × (5 -) = × (25-6) =. Solution: original ==+=+=-+-=-.5. Solution: Let x=n+2+, y=n+2-, then xy=2n+4, xy=4n+8. The original formula =+===-2=-2=n. When n=+1, the original formula =+. Solution: It is known that x=3+2, y=3-2, so x+y=6, xy=1, so the original formula ===. Solution: ======.8. Solution: original ====.9. Solution: From the property of the quadratic root, we get 3-5a=0, ∴a=.∴b=15, ∴a+b>0, a-b<0.∴-=-=-==. When a=, When b=15, the original formula =×=. Method dial: For algebraic expressions of the form ++2 or +-2, the form must be changed to or. When they are reduced as the squared number, a+b and a-b are required. And the symbol of ab. 10. Solution: Let x=k(k>0), then y=2k, z=3k, ∴ original ===-. Solution: ∵a+b=-8, ab=8, ∴a<0, b<0.∴b+a=--=-·=-=-=-=-12. Dial: Solve such questions and substitute for evaluation. You can't just think about the value of a single letter, which complicates the problem and can't even solve it. Decoding special training four 1.2 dial: from the title meaning 3x-4 = 0, x-y = 0, so x =, y = 4, substituting for evaluation. 2. Solution: From the absolute value and the non-negative property of the quadratic root, we get |2a-4|≥0, ≥0. And because 2|2a-4|+=0, the solution is a+b-ab=2-3- 2×(-3)=. ∵ ≥ 0, ∴ when 9x +1 = 0, that is, x = -, the value of +3 is the smallest, and the minimum value is 3. Method dialing: The minimum (large) value problem involving the quadratic root type is determined according to the specific situation of the problem. what way. In general, the non-negative solution of the quadratic root is used. 4. Solution: From the meaning of the problem, we know that a=0, substituting the known equation to get -=0, so =, so x=-y, so ===. Decoding the training five. Solution: AC===, ∵S△ABC=AC·BC=CD·AB, ∴CD===. Method law: use the area phase according to the nature of the right triangle. Solution: The upper base of the right angle trapezoid is 2cm, the lower base is cm, the height is cm, and the two waist lengths of the right angle trapezoid are cm, = (cm). The area of ??the trapezoid is (2) ×= (cm2). The circumference of the trapezoid is 2+++=5++(cm). Dial: This question examines the application of the quadratic root. The knowledge points used are the trapezoidal area formula and the Pythagorean theorem. The key to solving the problem is to master the trapezoidal area formula. 3. Solution: Over point B for BC⊥x axis intersection x axis at point C, as shown in the figure, by the meaning of the question, OA=, AB=2, ∵∠OBA=90°, ∴OB2=OA2-AB2=12-4= 8, the solution OB = 2. ∵ BC · OA = OB · BA, ∴ BC = =. In Rt △ OBC, OC = =, ∴ B point coordinates are (. 3) (question 4) 4. Solution: ∵A is 10m in the northeast direction of the water tower O, and B is 20m in the southeast direction of the water tower O. As shown in the figure, the plane rectangular coordinate system can be obtained to obtain A(5,5), B(20,-20). Point A is the vertical line of the x-axis, and point B is the perpendicular line of the y-axis. When the two perpendicular lines intersect at point C, AC=(20+5)m, BC=(20-5)m, according to the Pythagorean theorem, AB====30(m), that is, the length of the water pipe is 30m. A: The length of the water pipe is 30m. Dial: This question examines the application of the secondary root type, finds the coordinates of points A and B, and constructs the auxiliary line. A right triangle with AB as the hypotenuse is the key to solving the problem. Decoding training six 1. X≥-1. Solution: From the meaning of the question, get +=0, ∴x-3=0, y+2=0, solve x=3, y=-2, then 6x+y=16, the square root of ∴6x+y is ±. A. Solution: original = 5-1-9+-1-1+2=-7+. Solution: original = × = × + × = + =. When a =, b = -, the original = -. 8. Solution: ∵+|b-|=0, ∴a+1=0, b-=0, solution a=-1, b=. original =[-]÷=·=·===-.9. Solution: When the waist length is 2, the base length is 4+7-2×2=7, ∵2+2=4=<7, ∴ can not form a triangle at this time; when the base length is 2, the waist length is (4+7- 2) ÷ 2 = +, ∵ at this time the sum of any two sides is greater than the third side, ∴ can form a triangle. Comprehensive +.10. Solution: ∵DE=30m, AE=30÷=18(m), ∴AD===6(m)∵CD=10m, ∴EF=10m. ∵CF=DE=30m, FB=30÷=60 (m), ∴CB===30(m). ∴ circumference = AD + DC + CB + AB = 6 + 10 + 30 + 60 + 10 + 18 = 6 + 30 + 98 (m). Answer: The section perimeter of the dam is (6+30+98) Decoding Specialist VII 1. Solution: -=|a+2|-|a-1|, discussed in three cases: when a≤-2, the original formula = (-a-2)-[-(a-1)]=-a-2+a -1=-3; when -2 1, the original formula = (a + 2) - (a - 1) = 3 . Dial: When you want to find the sum or difference between two absolute values ??containing letters, you should classify the discussion. This problem can also determine the demarcation points by solving the inequality. 2. Solution: From the position of the corresponding point on the number axis of m, n, m>n, 00, m-1<0, n+1<0. =|m|+|n|+|m-n|+|n+1|-|m-1|=m-n+m-n-1-n-(1-m)=m-n+m-n-1-n -1+m=3m-3n-2. Method dialing: When using =|a| to simplify, we must combine the specific problem, first determine the symbol of the formula inside the absolute value, and then simplify. 3. Solution: (7+2-)(2-7+)=[2+(7-)][2-(7-)]=2)2-(7-)2=24-(98+3-14)=14-77. .

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