\(1\frac{1}{3}x-\left(\frac{3}{2}\right)^2=-0,75\)

\(\frac{4}{3}x-\frac{9}{4}=-\frac{3}{4}\)

\(\frac{4}{3}x=-\frac{3}{4}+\frac{9}{4}\)

\(\frac{4}{3}x=\frac{3}{2}\)

\(x=\frac{3}{2}:\frac{4}{3}\)

\(x=\frac{3}{2}.\frac{3}{4}\)

\(x=\frac{9}{8}\)

Vậy \(x=\frac{9}{8}\)

a) 2/3x:0.75=1

=> 2/3x=0.75

=> 2/3x=3/4

x= 3/4:2/3

x= 9/8

ok ko sao

2/3x : (-0.75)=1

2/3x=-0.75

2/3x=-3/4

x=-3/4:2/3

x=-9/8

-9/8 nhé

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

a: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{-1}{7}+\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)

\(\Leftrightarrow x=\dfrac{5}{14}:\dfrac{3}{5}=\dfrac{25}{42}\)

b: =>|3x-1|=2

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

c: \(\Leftrightarrow7\left(37-x\right)=3\left(x-13\right)\)

=>259-7x=3x-39

=>-10x=-298

hay x=29,8

d: =>x=3/4+2/3=9/12+8/12=17/12

2xX-12=8

2xX=8+12

2xX=20

x=20:2

X=10

0,75xX-1/4=1/2

0,75xX=1/2+1/4

0,75xX=3/4

X=3/4:0,75

X=1

1+2+3+4+..+9=3X(2+1)

(1+9)+(2+8)+(3+7)+(6+4)+5=3xXx3

45=9X

X=45:9

X=5

a, (2x-1)³=8

2x-1=2

2x=2+1

2x=3

x=1,5

a, \(\left(2x-1\right)^3=8\)

\(\Leftrightarrow\left(2x-1\right)^3=2^3\)

\(\Leftrightarrow2x-1=2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

b, \(\left(x-1\right)^x+2=\left(x-1\right)^x+4\)

\(\Leftrightarrow\left(x-1\right)^x+2-\left(x-1\right)^x-4=0\)

\(\Leftrightarrow-2\ne0\)=> vô nghiệm 

c, \(3^x-1+5.3^{3x}-1=162\)

Đề sai.

Bài 2: 

a: \(\left(3x-3\right)^2-\left(5x-3\right)^2=0\)

\(\Leftrightarrow\left(3x-3-5x+3\right)\left(3x-3+5x-3\right)=0\)

\(\Leftrightarrow-2x\left(8x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\)