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a: \(A=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: \(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5: \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\dfrac{8}{27}x^6-\dfrac{2}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^3-27}{x^2-9}\left(x\ne\pm3\right)\)
\(=\dfrac{x^3-3^3}{x^2-3^2}\)
\(=\dfrac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+3x+9}{x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3: \(\left(3x+5\right)\left(2x-7\right)\)
\(=6x^2-21x+10x-35\)
\(=6x^2-11x-35\)
4: \(\left(5x-2\right)\left(3x+4\right)\)
\(=15x^2+20x-6x-8\)
\(=15x^2+14x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3x-1}{x-1}-\frac{2x+5}{x-3}=1\)\(\left(ĐKXĐ:x\ne1;x\ne3\right)\)
\(\Leftrightarrow\)\(\frac{\left(x-3\right)\left(3x-1\right)}{\left(x-3\right)\left(x-1\right)}\)\(-\frac{\left(x-1\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)}\)\(=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}\)
\(\Rightarrow\)\(\left(x-3\right)\left(3x-1\right)-\left(x-1\right)\left(2x+5\right)=\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(3x^2-10x+3-2x^2-3x+5=x^2-4x+3\)
\(\Leftrightarrow\)\(x^2-13x+8=x^2-4x+3\)
\(\Leftrightarrow\)\(x^2-x^2-13x+4x=3-8\)
\(\Leftrightarrow\)\(-9x=-5\)
\(\Leftrightarrow\)\(x=\frac{5}{9}\)\(\text{(T/m ĐKXĐ)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\left(3x+2\right)\left(5-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)
\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)
\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{9}{46}\)
Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)
\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)
\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)
\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)
\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)
\(\Leftrightarrow6x-4=16\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
Vậy \(S=\left\{\dfrac{10}{3}\right\}\)
\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)
\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Rightarrow5x-5=4x+8\)
\(\Rightarrow x=13\left(tmdk\right)\)
Vậy \(S=\left\{13\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Câu 5: D