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21 tháng 7 2016

Ta có:\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(y+z\right)+\left(y+z\right)^2\)

\(=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2\)

\(=x^2\)

\(=x.x\)

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)

Cảm ơn bạn nhiều! Bạn có thể làm bài 1 không

 

27 tháng 7 2023

\(x^2+6x-7=0\\ \Leftrightarrow x^2-x+7x-7=0\\ \Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy \(S=\left\{1;-7\right\}\)

27 tháng 7 2023

\(x^2+6x-7=0\\ \Leftrightarrow x^2+7x-x-7=0\\ \Leftrightarrow\left(x^2+7x\right)-\left(x+7\right)=0\\ \Leftrightarrow x\left(x+7\right)-\left(x+7\right)=0\\ \Leftrightarrow\left(x+7\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

11 tháng 7 2017

Ta có:\(2\left(x-y\right)\left(z-y\right)+2\left(y-z\right)\left(z-x\right)+2\left(y-z\right)\left(x-z\right)\)

\(=2\left[\left(x-y\right)\left(z-y\right)+\left(y-x\right)\left(z-x\right)+\left(y-z\right)\left(x-z\right)\right]\)

\(=2\left[xz-xy-yz+y^2+yz-xy-zx+x^2+yx-yz-zx+z^2\right]\)

\(=2\left[-xz-xy-yz+x^2+y^2+z^2\right]\)

\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)

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

31 tháng 7 2017

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

(x+y+z)(x+z-y)(x+y-z)(y+z-x)

=[(x+y)^2-z^2]*[(x+z-y)(y+z-x)]

=[(x+y)^2-z^2][y^2-(x+z)^2]

=(x^2+2xy+y^2-z^2][y^2-x^2-2xz-z^2]

=x^2y^2-x^4-2x^3z-x^2z^2+2xy^3-2x^3y-4x^2yz-2xyz^2+y^4-y^2x^2-2xy^2z-z^2y^2-y^2z^2+x^2z^2+2xz^3+z^4

a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)

\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)

\(=x\left(x+2y+2z\right)\)

b: \(\left(x-3\right)^2-2\left(x^2-9\right)+\left(x+3\right)^2\)

\(=\left(x-3-x-3\right)^2\)

=36

c: \(\left(a^2-b^2\right)^2-\left(a+b^2\right)^2\)

\(=\left(a^2-b^2-a-b^2\right)\left(a^2-b^2+a+b^2\right)\)

\(=\left(a^2-a-2b^2\right)\left(a^2+a\right)\)

\(=a\cdot\left(a+1\right)\left(a^2-a-2b^2\right)\)

11 tháng 7 2018

a) Ta có:

(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2

=x^2 - y^2 - 2yz - z^2.

b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2

=x^2 +2xz- y^2 +z^2.

c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)

= -16 + x^2 - 6x + 9

= x^2 - 6x - 7.

11 tháng 7 2018

\(a,\left(x+y+z\right)\left(x-y-z\right)\)

\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)

\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)

\(=x^2-y^2-2yz-z^2\)

\(b,\left(x-y+z\right)\left(x+y+z\right)\)

\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)

\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)

\(=x^2+2xz-y^2+z^2\)

\(c,-16+\left(x-3\right)^2\)

\(=-16+x^2-6x+9\)

\(=x^2-6x-7\)

a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)

\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)

\(=\left(a^2\right)^2-\left(2a+3\right)^2\)

\(=a^4-\left(2a+3\right)^2\)

b: \(\left(-a^2-2a+3\right)^2\)

\(=\left(a^2+2a-3\right)^2\)

\(=a^4+4a^2+9+4a^3-18a-6a^2\)

\(=a^4+4a^3-2a^2-18a+9\)

c: \(\left(x-y-z\right)^2\)

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

\(=x^2-2xy-2xz+y^2+2yz+z^2\)

d: \(\left(x+y+z\right)\left(x-y-z\right)\)

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

\(=x^2-y^2-2yz-z^2\)