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\(8x^2-2\)
\(=2.\left(1^2-4x^2\right)\)
\(=2.\left(1-4x\right).\left(1+4x\right)\)
\(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right).\left(x-3+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-8y^2+6x+9\)
\(=\left(x^2+6x+9\right)-8y^2\)
\(=\left(x+3\right)^2-\left(\sqrt{8}\cdot y\right)^2\)
\(=\left(x+3+\sqrt{8}y\right)\left(x+3-\sqrt{8}y\right)\)
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Câu 3:
a: 2x-8=4
nên 2x=12
hay x=6
b: 7x-3x=2x+7
\(\Leftrightarrow4x-2x=7\)
hay \(x=\dfrac{7}{2}\)
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Câu 1:
a: \(5x\left(3x-4\right)=15x^2-20x\)
b: \(\left(x+5\right)\left(x-5\right)=x^2-25\)
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)
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\(\left(x-y\right)^5+\left(y-z\right)^5+\left(z-x\right)^5\)
\(=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5+y^5-5y^4z+10y^3z^2-10y^2z^3+5yz^4-z^5\)\(+z^5-5z^4x+10z^3x^2-10z^2x^3+5zx^4-x^5\)
\(=5\left(-x^4y+2x^3y^2-2x^2y^3+xy^4-y^4z+2y^3z^2-2y^2z^3+yz^4-z^4x+2z^3x^2-2z^2x^3+zx^4\right)\)
x3 - x2 - 3x + 27
= x3 + 3x2 - 4x2 - 12x + 9x + 27
= x2 ( x + 3 ) - 4x ( x + 3 ) + 9 ( x + 3 )
= ( x + 3 ) ( x2 - 4x + 9 )
x^3-x^2-3x+27
= (x+3)(x^2-4x+9)
nha bạn chúc bạn học tốt nha