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1, 4[3x + 4] = 12x + 16
2, 4/3[5/3x + 6] = 20/9x + 12
3, [4/3x + 7]. 5 + 2 = 20/3x + 35 + 2 = 20/3x + 37
4, [3/7x - 6][-7] + 3x = -3x + 42 + 3x = 42
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
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a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(3x-2\(^4\)).7\(^3\)=2.7\(^4\)
3x-2\(^4\)=2.7\(^4:7^3\)
3x-16=2.7
3x-16=14
3x=30
=>x=10
Vậy x=10
a) Ta có:\(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Leftrightarrow3x-16=2\cdot\dfrac{7^4}{7^3}=2\cdot7=14\)
\(\Leftrightarrow3x=30\)
hay x=10
Vậy: x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:\(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Leftrightarrow3x-16=2\cdot\dfrac{7^4}{7^3}=2\cdot7=14\)
\(\Leftrightarrow3x=30\)
hay x=10
Vậy: x=10
(3x - 24) . 73 = 2. 74
3x - 24 = 2.74 : 73
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3 = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
75 * (3x-23) = 74 * 72
3x -23 = 74 * 72 : 75
3x -23 = 7
3x = 7 - 8
3x = -1
x = \(\frac{-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4:
=>(2x+3,5)=7/12*3/14=21/168=1/8
=>2x=1/8-7/2=1/8-28/8=-27/8
=>x=-27/16
5: =>1/3:3x=-21/4
=>3x=-1/3:21/4=-1/3*4/21=-4/63
=>x=-4/189
6: =>2+7/9-3/4(x+1)=7/9
=>2-3/4(x+1)=0
=>3/4(x+1)=2
=>x+1=2:3/4=2*4/3=8/3
=>x=5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\frac{1}{2}+x=\frac{3}{2}-7\)
<=> \(\frac{3}{2}+x=\frac{-11}{2}\)
<=> \(x=-7\)
\(\frac{1}{4}+\frac{1}{3}:3x=-5\)
<=> \(\frac{1}{3}:3x=\frac{-21}{4}\)
<=> \(3x=\frac{-4}{63}\)
<=> \(x=\frac{4}{189}\)
\(\frac{4}{5}.x=\frac{8}{35}\)
<=> \(x=\frac{2}{7}\)
\(\frac{2}{3x}-\frac{1}{4}=\frac{7}{1}\)
<=> \(\frac{2}{3x}=\frac{29}{4}\)
=> \(8=87x\)
<=> \(x=\frac{8}{87}\)
\(\frac{3}{5x}+\frac{1}{2}=\frac{1}{7}\)
<=> \(\frac{3}{5x}=\frac{-5}{14}\)
<=> \(-25x=42\)
<=> \(x=\frac{-42}{25}\)
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16.\frac{2}{3}\right)=0\)
<=> \(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{-32}{3}=0\)
<=> \(\frac{43}{8}+x-\frac{173}{24}=\frac{-32}{3}\)
<=> \(\frac{43}{8}+x=\frac{-83}{24}\)
<=> \(x=\frac{-53}{6}\)
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