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Bài toán 1.2 :
\(3a\left(b^2-2c\right)-\left(a-4\right)\left(2c-b^2\right)\)
\(=3a\left(b^2-2c\right)+\left(a-4\right)\left(b^2-2c\right)\)
\(=\left(b^2-2c\right)\left(3a+a-4\right)\)
\(=\left(b^2-2c\right)\left(4a-4\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a)\(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b)\(x^2-5x+5y-y^2=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
a) (x-y)(x+y)-(x+y)=(x+y)(x-y-1) b) (x^2-y^2)-(5x-5y)=(x-y)(x+y)-5(x+y)=(x+y)(x-y-5) c) (x^3+1)-3x(x+1)=(x+1)(x^2-x+1)-3x(x+1)=(x+1)(x^2-4x+1)
\(Q=\left(x+2z\right)\left(3x^2+5x^2y\right)-\left(7x^2-3x^2y\right)\left(2z+x\right)\)
\(Q=3x^3+5x^3y+6x^2z+10x^2yz-14x^2z+6x^2yz-7x^3+3x^3y\)
\(Q=-4x^3+8x^3y+16x^2yz-8x^2z\)
\(Q=-4x^3\left(1-2y\right)+8x^2z\left(2y-1\right)\)
\(Q=4x^3\left(2y-1\right)+8x^2z\left(2y-1\right)\)
\(Q=\left(2y-1\right)\left(4x^3+8x^2z\right)\)