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\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)
2:
-8x^6-12x^4y-6x^2y^2-y^3
=-(8x^6+12x^4y+6x^2y^2+y^3)
=-(2x^2+y)^3
3:
=(1/3)^2-(2x-y)^2
=(1/3-2x+y)(1/3+2x-y)
Ta có:\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(y+z\right)+\left(y+z\right)^2\)
\(=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2\)
\(=x^2\)
\(=x.x\)
\(\frac{-x^6}{125}-\frac{y^3}{64}\)
\(=\frac{-\left(x^2\right)^3}{5^3}-\frac{y^3}{4^3}\)
\(=\left(\frac{-x^2}{5}\right)^3-\left(\frac{y}{4}\right)^3\)
\(=\left(\frac{-x^2}{5}-\frac{y}{4}\right)\cdot\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)
Tham khảo nhé~
Ta có:\(2\left(x-y\right)\left(z-y\right)+2\left(y-z\right)\left(z-x\right)+2\left(y-z\right)\left(x-z\right)\)
\(=2\left[\left(x-y\right)\left(z-y\right)+\left(y-x\right)\left(z-x\right)+\left(y-z\right)\left(x-z\right)\right]\)
\(=2\left[xz-xy-yz+y^2+yz-xy-zx+x^2+yx-yz-zx+z^2\right]\)
\(=2\left[-xz-xy-yz+x^2+y^2+z^2\right]\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)
\(=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
\(B=\left(\frac{x}{2}+y\right)^3-6\left(\frac{x}{2}+y\right)^2.z+6\left(x+2y\right)z^2-8z^3\)
\(=\left(\frac{x}{2}+y\right)^3-3.\left(\frac{x}{2}+y\right)^2.2z+3.\left(\frac{x}{2}+y\right).\left(2z\right)^2-\left(2z\right)^3\)
\(=\left(\frac{x}{2}+y-2z\right)^3\)
\(C=\left(m-n\right)^3+15\left(m-n\right)^2.\left(m-p\right)-75\left(n-m\right)\left(p-m\right)^2-125\left(p-m\right)^3\)
\(=\left(m-n\right)^3+3.\left(m-n\right).\left[5\left(m-p\right)\right]+3.\left(m-n\right).\left[5\left(m-p\right)\right]^2+\left[5\left(m-p\right)\right]^3\)
\(=\left(m-n+5m-5p\right)^3=\left(6m-n-5p\right)^3\)
làm tương tự
Viết các biểu thức sau dưới dạng tích các đa thức?
a)16x^2-9
b)9x^2-25y^2
c)49a^2-4y^4
d)8x^6-125y^6
e)(2x+y)^2-4
f)(x+y+z)^2-(x-y-z)^2
Bài làm
a)16x^2-9
=(4x)^2-3^2
=(4x-3)(4x+3)
b)9x^2-25y^2
=(3x)^2-(5y)^2
=(3x-5y)(3x+5y)
c)49a^2-4y^4
=(7a)^2-(2y^2)^2
=(7a-2y^2)(7a+2y^2)
d)8x^6-125y^6
=(2x^3)^3-(5y^3)^3
=(2x^3-5y^3)(2x^3+5y^3)
e)(2x+y)^2-4
=(2x+y-2)(2x+y+2)
f)(x+y+z)^2-(x-y-z)^2
=(x+y+z-x+y+z)(x+y+z+x-y-z)
=(2x+2y+2z)2x
\(a.9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
\(b.\left(2x+y\right)^2-1=\left(2x+y-1\right)\left(2x+y+1\right)\)
\(c.\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left[\left(x+y+z\right)+\left(x-y-z\right)\right]\left[\left(x+y+z\right)\right]-\left(x-y-z\right)\\ =2x.\left(2y+2z\right)\)
a) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
b) \(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y+1\right)\left(2x+y-1\right)\)
c) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left(x+y+z+x-y-z\right)\left(x+y+z-x+y+z\right)\)
\(=2x\left(2y+2z\right)\)