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1:
a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)
b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)
c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
b) \(y^3-13=\left(y-\sqrt{13}\right)\left(y^2+\sqrt{13}y+13\right)\)
c) \(2x^2-4=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\)
d) \(\left(x-1\right)^3-\left(y+1\right)^3=\left(x-1-y-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]=\left(x-y-2\right)\left(x^2-2x+1+xy-y+x-1+y^2+2y+1\right)=\left(x-y-2\right)\left(x^2+y^2-x+y+xy+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tích mình đi
Ai tích sẽ có lợi
vì khi có lợi bạn sẽ được người khác tích lại.
THANKS
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Đề bài không chính xác, biểu thức này không viết được dưới dạnh tích
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9/16x2m-2y2 - 2xmym + 16/9x2y2m-2
=\(\left(\dfrac{3}{4}x^{m-1}y\right)^2\) - 2xmym + \(\left(\dfrac{4}{3}xy^{m-1}\right)^2\)
=> \(\left(\dfrac{3}{4}x^{m-1}y-\dfrac{4}{3}xy^{m-1}\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(x-2)(x2+2x+4)-19=0
\(\Rightarrow\)x3-8-19=0
\(\Rightarrow\)x3-27=0
\(\Rightarrow\)(x-3)(x2+3x+9)=0
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+3\\x^2+3x=0-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+3x=-9\end{matrix}\right.\)
(6x-1)(6x+1)-15=0
\(\Rightarrow\)36x2-1-15=0
\(\Rightarrow\)36x2-16=0
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
b: Ta có: \(\left(x-3\right)^2-16\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x+1\right)\left(x-7\right)\)
c: \(y^2+16y+64=\left(y+8\right)^2\)
a) \(\left(x^2-2x-1\right)^2\)
\(=\left[x^2+\left(-2x\right)+\left(-1\right)\right]\left[x^2+\left(-2\right)+\left(-1\right)\right]\)
\(=\left(x^2\right)\left(x^2\right)+\left(x^2\right)\left(-2x\right)+\left(x^2\right)\left(-1\right)+\left(-2x\right)\left(x^2\right)+\left(-2x\right)\)
\(=x^4-2x^3-x^2-2x^3+4x^3+2x-x^2+2x+1\)
\(=x^4-4x^3+2x^2+4x+1\)
Mk ko chắc
a) \(\left(x^2-2x-1\right)^2\)
\(=\left(x^2-2x\right)^2-2\left(x^2+2x\right)-1\)
\(=x^4+4x^3-2x^2+4x^2+4x+1\)
\(=x^4+4x^3-2x^2+4x+1\)
b) Tương tự