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a) \(A_4=\left(x^2-3x+5\right)^2+7x\cdot\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)^2+4x\cdot\left(x^2-3x+5\right)+3x\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)\left(x^2-3x+5+4x\right)+3x\left(x^2-3x+5+4x\right)\)
\(=\left[\left(x^2-3x+5\right)+3x\right]\cdot\left(x^2-3x+5+4x\right)\)
\(=\left(x^2-3x+5+3x\right)\left(x^2+x+5\right)\)
\(=\left(x^2+5\right)\left(x^2+x+5\right)\)
\(A_5=2\left(x^2+5x-2\right)^2-7\left(x^2+5x-2\right)\left(x^3+3\right)+5\left(x^2+3\right)^2\)
Đặt \(x^2+5x-2=a;x^3+3=b\),Ta có:
\(2a^2-7ab+5b^2=2a^2-5ab-2ab+5b^2=a\left(2a-5b\right)-b\left(2a-5b\right)=\left(2a+5b\right)\left(a-b\right)\)
Thay \(x^2+5x-2=a;x^3+3=b\),ta có:
.......................
bn làm nốt nhé
a) A = (x2 - 2.x.3 + 32) - (3y)2
A = (x - 3)2 - (3y)2
A = (x - 3 - 3y)(x-3+3y)
b) B = (x-1)3 + 2(x-1)(x+1)
B=(x-1)[(x-1)2 - 2(x+1)]
B = (x-1)[x2 - 4x - 3]
1.
7x(2x-1)=14x2-7x
2
a. x2+2x=x(x+2)
b.x2-xy+3x-3y
=x(x-y)+3(x-y)
=(x+3)(x-y)
Câu 2:
1. 2x/2x-5 - 5/2x-5
=2x-5/2x-5
=1
2. (6x3-7x2-x+2) : (x-1)=6x2-x-2
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
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)
e/=\(2x^2-2x+5x-5=2x\left(x-1\right)-5\left(x-1\right)\)
=\(\left(2x-5\right)\left(x-1\right)\)
f/=\(x^2-4xy+4y^2-3xy+6y^2\)
=\(\left(x-2y\right)^2-3y\left(x-2y\right)\)=\(\left(x-2y\right)\left(x-2y-3y\right)\)
=\(\left(x-2y\right)\left(x-5y\right)\)
a/ = \(\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
=\(\left(2x-5\right)\left(2x+5-2x-7\right)\)=\(\left(2x+5\right)\left(-2\right)\)
b/=\(\left(x+2\right)^2-y^2\)=\(\left(x+2+y\right)\left(x+2-y\right)\)
c/=\(\left(x-1\right)^3\)
d/=\(x^2-5x+3x-15=x\left(x-5\right)-3\left(x-5\right)\)
=\(\left(x-5\right)\left(x-3\right)\)
\(A_3=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(t=x^2+4x+10\)
\(A_3=t^2-7\left(t^2+1\right)+7\)
\(=-6t^2\)
Thay vào : \(-6\left(x^2+4x+10\right)^2\)
2 , \(A_1=\left(t^2+3x\right)^2-2\left(x^2+3x\right)-8\)
Đặt \(t=x^2-3x\)
\(A_1=t^2-2x-8=\left(t-4\right)\left(t+2\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)