Tìm x:

a)

b)

c) 

x=\(\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

a) x + 4/5 = 3/2

x = 3/2 - 4/5

x = 7/10

b) x - 3/2 = 11/4

x = 11/4 + 3/2

x = 17/4

c) 25/3 - x = 5/6

x = 25/3 - 5/6

x = 15/2

\(35x^2-49=10x^2< =>25x^2-7^2=0\)

\(< =>\left(5x\right)^2-7^2=0< =>\left(5x+7\right)\left(5x-7\right)=0\)

\(=>\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=\dfrac{7}{5}\end{matrix}\right.\)

Ta có: \(35x^2-49=10x^2\)

\(\Leftrightarrow25x^2-49=0\)

\(\Leftrightarrow\left(5x-7\right)\left(5x+7\right)=0\)

\(\Leftrightarrow x\in\left\{\dfrac{7}{5};-\dfrac{7}{5}\right\}\)

\(\text{a, (x - 1)^2 = 1}\\ x-1=1\\ x=2\)

\(x-1=1\Rightarrow x=2\)

a)\(1,5-2\left|x\right|=-0,5\)

\(\Leftrightarrow2\left|x\right|=1,5+0,5\)

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

x      = 9/3 - 5/8

x      =  57/24

x+5/8=9/3

x       -= 9/3 - 5/8

x        =  57/24

\(x+\dfrac{5}{8}=\dfrac{9}{3}\)

\(x=\dfrac{9}{3}-\dfrac{5}{8}=\dfrac{19}{8}\)

\(\left[2040:\left(47-3x\right)+33\right].11=1683\)

\(\Leftrightarrow2040:\left(47-3x\right)+33=153\)

\(\Leftrightarrow2040\left(47-3x\right)=120\)

\(\Leftrightarrow47-3x=17\)

\(\Leftrightarrow3x=30\Leftrightarrow x=10\)

\(x+\dfrac{2}{-15}=\dfrac{5}{3}\)

\(x=\dfrac{9}{5}\)

\(x+\dfrac{2}{-15}=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}-\dfrac{-2}{15}\)

\(x=\dfrac{-23}{15}\)