\(x-\dfrac{3}{17}(2x-1)=\dfrac{7}{34}(1-2x)+\dfrac{10x-3}{2}\)
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a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
![](https://rs.olm.vn/images/avt/0.png?1311)
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(-4\le\dfrac{x^2-2x-7}{x^2+1}\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x-7\le x^2+1\\-4x^2-4\le x^2-2x-7\end{matrix}\right.\) (Do \(x^2+1>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\\left[{}\begin{matrix}x\ge1\\x\le-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\-4\le x\le-\dfrac{3}{5}\end{matrix}\right.\)
2.
\(\dfrac{1}{13}\le\dfrac{x^2-2x-2}{x^2-5x+7}\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+7\le13x^2-26x-26\\x^2-2x-2\le x^2-5x+7\end{matrix}\right.\) (Do \(x^2-5x+7>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{11}{4}\\x\le-1\end{matrix}\right.\\x\le3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{4}\le x\le3\\x\le-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
Lời giải:
PT $\Leftrightarrow \frac{11}{17}x+\frac{3}{17}=\frac{78}{17}x-\frac{22}{17}$
$\Leftrightarrow \frac{25}{17}=\frac{67}{17}x$
$x=\frac{25}{67}$
\(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
<=>\(x-\dfrac{3}{17}\left(2x-1\right)-\dfrac{7}{34}\left(1-2x\right)=\dfrac{10x-3}{2}\)
<=>\(x-\dfrac{3}{17}\left(2x-1\right)+\dfrac{7}{34}\left(2x-1\right)=\dfrac{10x-3}{2}\)
<=>\(x+\left(-\dfrac{3}{17}+\dfrac{7}{34}\right)\left(2x-1\right)-\dfrac{10x-3}{2}=0\)
<=>\(x+\dfrac{1}{34}\left(2x-1\right)-\dfrac{10x-3}{2}=0\)
<=>\(x+\dfrac{1}{17}x-\dfrac{1}{34}-\dfrac{10x-3}{2}=0\)
<=>\(\dfrac{18}{17}x-\dfrac{85x-25.5}{17}=\dfrac{0.5}{17}\)
=> \(18x-85x+25.5=0.5\)
<=>\(-67x=0.5-25.5\)
<=>\(-67x=-25\)
<=> \(x=\dfrac{25}{67}\)
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