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Hãy ôn lại phần:Pương chình dạng tích - Toán lớp 8 - sách giáo khoa
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8y^3 - 12y^2 + 6y + 1 - 2y(4y^2 - 12y + 9) - 12y^2 + 12y
= 8y^3 - 12y^2 + 6y + 1 - 8y^3 + 24y^2 - 18y - 12y^2 + 12y
= 1
=> đpcm
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\(a,\left(12x^3y^3-3x^2y^3+4x^2y^4\right):6x^2y^3\)
\(=12x^3y^3:6x^2y^3-3x^2y^3:6x^2y^3+4x^2y^4:6x^2y^3\)
\(=2x-\frac{1}{2}+\frac{2}{3}y\)
\(b,\left(6x^3-19x^2+23x-12\right):\left(2x-3\right)\)
\(=\left(6x^3-10x^2+8x-9x^2+15x-12\right):\left(2x-3\right)\)
\(=\left(2x-3\right)\left(3x^2-5x+4\right):\left(2x-3\right)\)
\(=3x^2-5x+4\)
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ta có 4 x 3 y 2 – 8 x 2 y 3 = 4 x 2 y 2 . x – 4 x 2 y 2 . 2 y = 4 x 2 y 2 ( x – 2 y )
Vậy 4x3y2 – 8x2y3 = 4x2y2(x – 2y)
Đáp án cần chọn là: C
bấm đúng cho mik đi
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\(A=\left(x-2\right)^2-\left(2x+1\right)^2=x^2-4x+4-4x^2-4x-1=-3x^2+3=-3\left(x^2-1\right)\)
\(=-3\left(x-1\right)\left(x+1\right)\)
\(B=\left(x-2y\right)^2-\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(x-2y-x-2y\right)=-4y\left(x-2y\right)\)
\(C=\left(x+1\right)^3-\left(x-2\right)^3=\left(x^3+3x^2+3x+1\right)-\left(x^3-6x^2+12x-8\right)\)
\(=x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-9x+9=9\left(x^2-x+1\right)\)
\(D=\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2=\left(x-1-x-1\right)^2=-2^2=4\)
\(E=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+2y-x=x^2+4xy+4y^2+2\left(x^2-4y^2\right)+2y-x\)
\(=x^2+4xy+4y^2+2x^2-8y^2+2y-x=3x^2-4y^2+4xy+2y-x\)
\(G=\left(2x+1\right)^3-\left(2x-1\right)=8x^3+12x^2+6x+1-2x+1=8x^3+12x^2+4x+2\)
\(=2\left(4x^3+6x^2+2x+1\right)=2\left(4x\left(x+1\right)^2+1\right)\)
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đề dài nên T giải câu a thôi bn tự làm tiếp mấy câu khác nhé
2x^2 - 2y^2 - 6x - 6y
= 2(x^2-y^2) - 6(x+ y)
= 2(x-y)(x+y) - 6(x+y)
= (2(x-y)-6) (x+y)
\(=\left(3-2y\right)^2\) (hđt số 2)
\(=3^2-2.2y+\left(2y\right)^2\)
\(=9-4y+4y^2\)