4X & # 178; - (X-Y) &# 178; factorization, I hope someone else will answer, quick 4x²-（x-y）² x²-（x-y）² -（a+b）²+（2a-3b）² 49（x-2）²-25（x-3）²

4X & # 178; - (X-Y) &# 178; factorization, I hope someone else will answer, quick 4x²-（x-y）² x²-（x-y）² -（a+b）²+（2a-3b）² 49（x-2）²-25（x-3）²

4x²-（x-y）²=(2x+x-y)(2x-x+y)=(3x-y)(x+y)x²-（x-y）²=(x+x-y)(x-x+y)=y(2x-y)-（a+b）²+（2a-3b）² =(2a-3b+a+b)(2a-3b-a-b)=(3a-2b)(a-4b)49（x-2）²-25（x-3）²=(7x-1...
3 / 2-1 / 3x-1 = 3 / 6x-2 factorization answer
Multiply both sides by 2 (3x-1)
3(3x-1)-2=3
9x-3-2=3
9x=8
x=8/9
The test shows that x = 8 / 9 is the solution of the equation
Read and solve the problem. For the quadratic trinomial like x2 + 2aX + A2, we can decompose it into (x + a) 2 by formula method. But for the quadratic trinomial x2 + 2ax-3a2, we can't directly use the formula. At this time, we can add a term A2 to the quadratic trinomial x2 + 2ax-3a2 to make the sum of it and X2 + 2aX a complete square, and then subtract the sum of A2 and the whole formula In this way, the method of adding an appropriate term to make a complete square expression appear in the expression, and then subtracting this term to make the value of the whole expression unchanged is called "matching method." (1) use the "matching method" to decompose the factor: a2-6a + 8. (2) if a + B = 5, ab = 6, find: ① A2 + B2; ② A4 + B4 (3) given that x is a real number, compare the size of x2-4x + 5 and - x2 + 4x-4, and explain the reason
(1) (2) = (2) = (2) (- 2), + 2) = (2) (- 2), + 2) = (2) (- 2), + 2); (2) = (2) = (2) (- 2), + 2) = (2) = (2) (- 2), + 2) = (2) (- 2), + 2); (2) = (2) = (2) (- 2), + 2) = (2) (- 3); (a) = (2) = (2) = (2) = (2) = (2) (- 2), + 2) = (2); (2) = (2) (- 2) = (2) (- 2), + 2) = (2); (2) = (2) = (2) (- 2) (- 2
The solution equation (1) (3x-1) (X-2) = (4x + 1) (X-2) (2) 5x2 − 4x − 14 = x2 + 34x
(1) (3x-1) (X-2) = (4x + 1) (X-2) transfer: (3x-1) (X-2) - (4x + 1) (X-2) = 0, decompose factor: (X-2) (3x-1-4x-1) = 0, get X-2 = 0 or - X-2 = 0, solution: X1 = 2, x2 = - 2; (2) remove denominator: 20x2-16x-1 = 4x2 + 3x, sort out: 16x2-19x-1 = 0, where a = 16, B = - 19, C = - 1, ∫ △ = 192 + 64 = 235 ＞ 0, ∫ x = 19 ± 23532, then X1 = 19 + 23532, X = 19 2=19−23532．
How to factorize 3x ^ 2 + 6x + 3?
Reflect the method
3x^2+6x+3
=3(x²+2x+1)
=3(x+1)²
For the quadratic trinomial x ^ 2 + 2aX + A ^ 2, we can use the formula method to decompose it into the form of (x + a) 2
(x + a) ^ 2, but for the quadratic trinomial x ^ 2 + 2ax-3a ^ 2, we can use the following method: x ^ 2 + 2ax-3a ^ 2 = x ^ 2 + 2aX + A ^ 2-A ^ 2-3a ^ 2 = (x + a) ^ 2 - (2a) ^ 2 = (x + 3a) (x-a), use the above method to decompose the following formula: (x + 1) ^ 4 + (x ^ 2-1) ^ 2 + (x-1) ^ 4, (m ^ 2-1) (n ^ 2-1) + 4Mn
(x+1)^4+(x^2-1)^2+(x-1)^4=(x+1)^4+2(x^2-1)^2+(x-1)^4-(x^2-1)^2=[(x+1)^2+(x-1)^2]^2-(x^2-1)^2 =(2x^2+2)^2-(x^2-1)^2=(3x^2+1)(x^2+3)(m^2-1)(n^2-1)+4mn=m^2*n^2-m^2-n^2+1+2mn+2mn=(mn+1)^2-（m-n）^2=(mn+m-...
Find out (X & # 178; - 4 / X & # 178; - 4x + 4 + 2-x / x + 2) divided by X / X-2
How to transform x + 2 of 2-x
[(x²-4)/(x²-4x+4) +(2-x)/(x+2)]/[x/(x-2)]=[(x+2)(x-2)/(x-2)² -(x-2)/(x+2)]/[x/(x-2)]=[(x+2)/(x-2)-(x-2)/(x+2)]/[x/(x-2)]=[(x+2)²-(x-2)²](x-2)/[x(x-2)(x+2)]=8x(x-2)/[x(x-2)(x+2...
Factorization: 3x ^ 3 + 6x ^ 2=_ (x+2)“_” What are you filling in?
3x^3+6x^2
=3x^2(x+2)
So fill in "3x ^ 2"
3 x^2
3X^2
3x^3+6x^2=3(x+2)“x²”
Find the common factor and put forward 3x ^ 2
How to use the sup2; & sup2; - AB + 2 formula to solve the equation
Wrong, X & sup2; - 2ax-b & sup2; + A & sup2; = 0
Just use the root formula, - B & sup2; + A & sup2; instead of C in the formula
have to
x1=a+b or x2=a-b
Just simplify the formula again ~ ~ ~ the formula is too complex, it's not easy to send on the Internet...
x^2-2ax+a^2-b^2=0
(x-a)^2-b^2=0
(x-a+b)(x-a-b)=0
x1=a-b..x2=a+b
What is the solution of X & # 178; = 4x?
X = 0 or x = 4
Thank you for discussing and studying with each other