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a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left(x^2-3\right)^2=0\\ \Leftrightarrow x^2-3=0\\ \Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\\ b,\Leftrightarrow8x^3+12x^2+6x+1-64=0\\ \Leftrightarrow\left(2x+1\right)^3-4^3=0\\ \Leftrightarrow\left(2x+1-4\right)\left[\left(2x+1\right)^2+4\left(2x+1\right)+16\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=3\\4x^2+4x+1+8x+4+16=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\4x^2+12x+17=0\left(1\right)\end{matrix}\right.\)
Xét \(\left(1\right)\Leftrightarrow\left(2x+3\right)^2+8=0\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
Vậy pt có nghiệm \(x=\dfrac{3}{2}\)
\(c,\Leftrightarrow\left(3-2x-5\right)\left(3-2x+5\right)=0\\ \Leftrightarrow\left(-2-2x\right)\left(8-2x\right)=0\\ \Leftrightarrow-2\left(x+1\right)\cdot2\left(4-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow25x=-25\Rightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=x^3-3x^2y+3xy^2-y^3-3x^3+6x^2y-3xy^2+3x^3-3x^2y-x^3\\ D=-y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Pt $\Leftrightarrow (x-4)^3=0\\\Leftrightarrow x-4=0\\\Leftrightarrow x=4$
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\(\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
\(=\left(5x+3y\right)\left[\left(5x\right)^2-5x.3y+\left(3y\right)^2\right]\)
\(=\left(5x\right)^3+\left(3y\right)^3=125x^3-27y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)=6a^2b+b^3=b\left(6a^2+b^2\right)\)
b) \(\left(x+y\right)^3+\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)+\left(x^3-3x^2y+3xy^2-y^3\right)=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)
a) \(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
b) \(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\left(x^2+3y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3a^2b+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
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