Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(-9x^2+12xy-4y^2\)
\(=-\left(9x^2-12xy+4y^2\right)\)
\(=-\left[\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\right]\)
\(=-\left(3x-2y\right)^2\)
b) Ta có: \(-125a^3+75a^2-15a+1\)
\(=\left(-5a\right)^3+3\cdot\left(-5a\right)^2\cdot1+3\cdot\left(-5a\right)\cdot1^2+1^3\)
\(=\left(-5a+1\right)^3\)
\(=\left(1-5a\right)^3\)
c) Ta có: \(64-96a+48a^2-8a^3\)
\(=4^3-3\cdot4^2\cdot2a+3\cdot4\cdot\left(2a\right)^2-\left(2a\right)^3\)
\(=\left(4-2a\right)^3\)
\(=\left[2\cdot\left(2-a\right)\right]^3\)
\(=8\left(2-a\right)^3\)
d) Ta có: \(-\frac{1}{8}m^3n^6-\frac{1}{27}\)
\(=-\left(\frac{1}{8}m^3n^6+\frac{1}{27}\right)\)
\(=-\left[\left(\frac{1}{2}mn^2\right)^3+\left(\frac{1}{3}\right)^3\right]\)
\(=-\left(\frac{1}{2}mn^2+\frac{1}{3}\right)\left(\frac{1}{4}m^2n^4-\frac{1}{6}mn^2+\frac{1}{9}\right)\)
\(12xy-4x^2y+8xy^2\)
\(=4xy\left(3-x+2y\right)\)
\(-125a^3+75a^2-15a+1\)
\(=-\left(125a^3-75a^2+15a-1\right)\)
\(=-\left[\left(5a\right)^3-3.\left(5a\right)^2.1+3.5a.1^3-1^3\right]\)
\(=-\left(5a-1\right)^3\)
a) \(16x^2-\left(x^2+4\right)^2=\left(4x\right)^2-\left(x^2+4\right)^2\)
\(=\left(4x+x^2+4\right)\left(4x-x^2-4\right)\)
\(=\left(x+2\right)^2\left\{-\left(x^2-4x+4\right)\right\}=\left(x+2\right)^2\left\{-\left(x-2\right)^2\right\}\)
Ở đây mình không đổi \(-\left(x-2\right)^2=\left(2-x\right)^2\)được vì vốn dĩ \(\left(x-2\right)^2=\left(2-x\right)^2\)
b) \(\left(x^2+9\right)^2-36=\left(x^2+9\right)^2-6^2\)
\(=\left(x^2+9+6\right)\left(x^2+9-6\right)=\left(x^2+15\right)\left(x^2-3\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
f) \(x^4y^4-z^4=\left\{\left(x^2y^2\right)^2-\left(z^2\right)^2\right\}\)
\(=\left\{\left(xy\right)^2-z^2\right\}\left\{x^2y^2+z^2\right\}\)
\(=\left(xy-z\right)\left(xy+z\right)\left(x^2y^2+x^2\right)\)
a) \(x^4-9x^2\)
\(=x^2\left(x^2-9\right)\)
\(=x^2\left(x-3\right)\left(^{ }x+3\right)\)
b) \(3x^2-12x+12\)
\(=3x\left(x^2-4x+4\right)\)
\(=3x\left(x-2\right)^2\)
c) \(x^2+5x+6\)
\(=x^2+3x+2x+6\)
\(=x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
x4 - 9x2
= x4 - ( 3x )2
= ( x2 - 3x ) ( x2 + 3x )
b) 3x3 - 12x2 + 12x
= 3x3 - 6x2 - 6x2 + 12x
= 3x2( x - 2 ) - 6x ( x - 2 )
= ( 3x2 - 6x ) ( x - 2 )
= 3x ( x - 2 ) ( x - 2 )
= 3x ( x- 2 )2
c) x2 + 5x + 6
= x2 + 2x + 3x + 6
= x ( x + 2 ) + 3 ( x + 2 )
= ( x + 3 ) ( x + 2 )
\(a)\) \(3x^2-6x=3x\left(x-2\right)\)
\(b)\) \(9x^3-9x^2y-4x+4y\)
\(=9x^2.\left(x-y\right)-4\left(x-y\right)\)
\(=\left(9x^2-4\right)\left(x-y\right)\)
\(=[\left(3x\right)^2-2^2]\left(x-y\right)\)
\(=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)
\(c)\) \(x^3-2x^2-8x\)
\(=x\left(x^2-2x-8\right)\)
\(=x\left(x+2\right)\left(x-4\right)\)
a) 25 - x2 + 4xy - 4y2 = 25 - (x2 - 4xy + 4y2) = 52 - (x - 2y)2 = (5 + x - 2y)(5 - x +2y) = (x - 2y + 5)(2y - x + 5)
b) 3a2c2 + bd + 3abc + acd = (3a2c2 + 3abc) + (bd + acd) = 3ac(ac + b) + d (ac + b) = (ac + b)(3ac + d)
c) x3 - 2x2 - x + 2 = x2(x - 2) - (x - 2) = (x - 2)(x2 - 1) = (x - 2)(x - 1)(x + 1)
d) a4 + 5a3 + 15a - 9 = (a4 + 3a2) + (5a3 + 15a) - (3a2 + 9) = a2(a2 + 3) + 5a(a2 + 3) - 3(a2 + 3) = (a2 + 3)(a2 + 5a - 3)
giúp em với ạ TvT em đang cần gấp lắm huhu
b)
Sửa đề: \(125a^3+75a^2+15a+1\)
Ta có: \(125a^3+75a^2+15a+1\)
\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)
\(=\left(5a+1\right)^3\)
c) Ta có: \(64-96a+48a^2-8a^3\)
\(=-\left(8a^3-48a^2+96a-64\right)\)
\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)
\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)
\(=-8\left(a-2\right)\left(a^2-4a+4\right)\)
\(=-8\left(a-2\right)^3\)