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Bài giải
a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)
\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)
\(\frac{2}{7}x=-\frac{5}{4}\)
\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)
\(x=-\frac{35}{8}\)
b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)
\(6x=-\frac{8}{125}-\frac{2}{5}\)
\(6x=-\frac{58}{125}\)
\(x=-\frac{58}{125}\text{ : }6\)
\(x=\frac{-29}{375}\)
c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\ne2\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)
\(\Rightarrow x^2-4=5\Rightarrow x^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne-1\)
\(\Rightarrow\left(x+1\right)^2=2.8=16\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
c) giống câu a
d) ĐKXĐ: \(x\ne5,x\ne-1\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x-5\right)\)
\(\Rightarrow x^2+3x+2=x^2-8x+15\)
\(\Rightarrow11x=13\)
\(\Rightarrow x=\dfrac{13}{11}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x=-213:(1+2+3+4+...+100)<=>x=-213/100
b) x-x=-1/3-2/4 <=> 0= -5/6 (vô lý )
c) x=-0,8119408369
d) x= 0.0258907758
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
(1/3.x-2/3)^8 hay ^3 vậy bạn?
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