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a ) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow-2\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\Leftrightarrow x=\dfrac{5}{2}.\)
Vậy .........
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-4\end{matrix}\right.\)
Vậy .........
c ) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy .........
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a: \(=x^3-27-x^3-27x+x+27=-26x\)
b: \(=x^2-14x-10x^2+20x-10=-9x^2+6x-10\)
c: \(\Leftrightarrow2x^2-4x-4x^2-6x+2x+3=0\)
=>3=0(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2x^3+x^2+x+6}{x^2-x+2}=\frac{\left(2x+3\right)\left(x^2-x+2\right)}{x^2-x+2}=2x+3\)
b) \(\frac{x}{x-3}-\frac{5x^2+27}{x^2-9}+\frac{x-9}{x+3}\)
\(=\frac{x}{x-3}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x-9}{x+3}\)
\(=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-9\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x-\left(5x^2+27\right)+x^3-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x^2-9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x}{x-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
=>(x-2)(x^2+4x+6)=0
=>x-2=0
=>x=2
b: =>(2x-5)(2x+5)-(2x-5)(2x+7)=0
=>(2x-5)(2x+5-2x-7)=0
=>2x-5=0
=>x=5/2
c: =>(x+3)(x^2-3x+9+x-9)=0
=>(x+3)(x^2-2x)=0
=>\(x\in\left\{0;2;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) = (x-3)(x+3) +(x-3((x-3)
= (x-3)(x+3+x-3)
= 2x(x-3)
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