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N = 8x3 – 12x2y + 6xy2 – y3
= (2x)3 – 3(2x)2y + 3.2xy2 – y3 (Hằng đẳng thức (5))
= (2x – y)3
Thay x = 6, y = - 8 ta được:
N = (2.6 – (-8))3 = 203 = 8000
Ta có
A = 8 x 3 – 12 x 2 y + 6 x y 2 – y 3 + 12 x 2 – 12 x y + 3 y 2 + 6 x – 3 y + 11 = ( 2 x ) 3 – 3 . ( 2 x ) 2 . y + 3 . 2 x . y - y 3 + 3 ( 4 x 2 – 4 x y + y 2 ) + 3 ( 2 x – y ) + 11 = ( 2 x – y ) 3 + 3 ( 2 x – y ) 2 + 3 ( 2 x – y ) + 1 + 10 = ( 2 x – y + 1 ) 3 + 10
Thay 2x – y = 9 vào A = ( 2 x – y + 1 ) 3 + 10 ta được
A = ( 9 + 1 ) 3 + 10 = 1010
Vậy A = 1010
Đáp án cần chọn là: C
\(-8x^3+12x^2y-6xy^2+y^3=\left(-2x\right)^3+3.\left(-2x\right)^3y+3.\left(-2x\right).y^2+y^3\)
\(=\left(-2x+y\right)^3\) (hay \(\left(y-2x\right)^3\) tùy cách ghi)
Ta có: \(-8x^3+12x^2y-6xy^2+y^3\)
\(=-\left(8x^3-12x^2y+6xy^2-y^3\right)\)
\(=-\left(2x-y\right)^3\)
Ta có
8 x 3 – 12 x 2 y + 6 x y 2 – y 3 = ( 2 x ) 3 – 3 . ( 2 x ) 2 y + 3 . 2 x . y 2 – y 3 = ( 2 x – y ) 3
Đáp án cần chọn là: A
a: \(A=\left(x+2y\right)^3=\left(-5\right)^3=-125\)
b: \(B=\left(2x-y\right)^3=\dfrac{1}{125}\)
c: \(=x^3+3x^2+3x+1-x^3+3x^2-3x+1-3x\left(x^2-2x+1+x+1\right)\)
\(=6x^2+2-3x\left(2x^2-x+2\right)\)
\(=6x^2+2-6x^3+3x^2-6x\)
\(=-6x^3+9x^2-6x+2\)
a.
\(M=x^2+4y^2-4xy=\left(x-2y\right)^2=\left(18-4.2\right)^2=10^2=100\)
b.
\(N=\left(2x\right)^3-3.\left(2x\right)^2.y-3.2x.y^2-y^3=\left(2x-y\right)^3=\left(2.6-\left(-8\right)\right)^3\)
\(=20^3=8000\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
1)
2x^2 + 12 + 2x.(4-x) = 0
2x^2 + 12 + 8x - 2x^2 = 0
12 + 8x = 0
8x = -12
x = -4/3
2)
A= 8x3 - 12x2y +6xy2 - y3
A = (2x)3 - 3.(2x)2.y + 3.2.x.y2 - y3
A = (2x-y)3
...
tính giá trị mk nhường bn làm đó!