Câu hỏi của Kaito Kun

Question

Viet cac da thuc sau duoi dang tich :

a, 100x2−(x2+25)2

b, 1+(x−y+5)2−2(x−y+5)

c, (x2+4y2−5)2−16(x2y2+2xy+1)

d, (x2+8x−34)2−(3x2−8x−2)2

Yukru · Accepted Answer

a) $100x^2-\left(x^2+25\right)^2$

$=\left(10x\right)^2-\left(x^2+25\right)^2$

$=\left(10x+x^2+25\right)\left(10x-x^2-25\right)$

$=\left(x+5\right)^2\left(10x-x^2-25\right)$

b) $1+\left(x-y+5\right)^2-2\left(x-y+5\right)$

$=\left[\left(x-y+5\right)-1\right]^2$

$=\left(x-y+4\right)^2$

c) $\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)$

$=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2$

$=\left[\left(x^2+4y^2-5\right)-4\left(xy+1\right)\right]\left[\left(x^2+4y^2-5\right)+4\left(xy+1\right)\right]$

$=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)$

$=\left(x^2-4xy+4y^2-9\right)\left(x^2+4xy+4y^2-1\right)$

$=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1^2\right]$

$=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)$

d) $\left(x^2+8x-34\right)^2-\left(3x^2-8x-2\right)^2$

$=\left[\left(x^2+8x-34\right)-\left(3x^2-8x-2\right)\right]\left[\left(x^2+8x-34\right)+\left(3x^2-8x-2\right)\right]$

$=\left(x^2+8x-34-3x^2+8x+2\right)\left(x^2+8x-34+3x^2-8x-2\right)$

$=\left(-2x^2+16x-32\right)\left(4x^2-36\right)$

$=-2\left(x^2-8x+16\right)\left[\left(2x\right)^2-6^2\right]$

$=-2\left(x-4\right)^2\left(2x-6\right)\left(2x+6\right)$

$=-2\left(x-4\right)^24\left(x-3\right)\left(x+3\right)$

$=8\left(x-4\right)^2\left(x-3\right)\left(x+3\right)$Câu trả lời: a) $100x^2-\left(x^2+25\right)^2$