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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)
Ta có:
x4+2x3+x2+x+1=(x2)2+2.x2.x+x2+x+1
=(x2+x)+(x+1)
=x2+2x+1
=(x+1)2
dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)+64x^4\)
=\(\left[\left(x^2-2x+2\right)^4+2.10x^2\left(x^2-2x+2\right)^2+100x^4\right]\)-100x4+64x2
=\(\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^2\)
=\(\left[\left(x^2-2x+2\right)^2+4x^2\right].\left[\left(x^2-2x+2\right)^2+16x^2\right]\)
1. \(x\left(x^2-5xy-14y^2\right)=x\left(x^2-7xy+2xy-14y^2\right)\)
\(=x\left(x-2\right)\left(x-7\right)\)
2. \(x^4+2x^2+1-9x^2=\left(x^2+1\right)^2-\left(3x\right)^2=\left(x^2+1-3x\right)\left(x^2+1+3x\right)\)
3. \(4x^4+4x^2+1-16x^2=\left(2x^2+1\right)^2-\left(4x\right)^2=\left(2x^2-4x+1\right)\left(2x^2+4x+1\right)\)
4. \(x^2+x+7x+7=\left(x+7\right)\left(x+1\right)\)
5. \(x\left(x^2-5x-14\right)=x\left(x^2-7x+2x-14\right)=x\left(x+2\right)\left(x-7\right)\)
Phân tích đa thức thành nhân tử :
1.x3-5x2y-14xy2
2.x4-7x2+1
3.4x4-12x2+1
4.x2+8x+7
5.x3-5x2-14x
bậc to thế ==
ừ