a, 1,5 +|2x - 2/3| = 3/2

            |2x - 2/3| = 3/2 - 1,5 

            |2x - 2/3| = 0

<=> 2x - 2/3 = 0 

<=> 2x = 0 + 2/3

<=> 2x = 2/3

<=> x = 2/3 : 2

<=> x = 1/3 

Vậy x = 1/3

b, 3/4 - |1/4 - x| = 5/8

            |1/4 - x| = 3/4 - 5/8

            |1/4 - x| = 1/8

<=> 1/4 - x = 1/8

       1/4 - x = /1/8

<=> x = 1/4 - 1/8

       x = 1/4 - ( -1/8)

<=> x = 1/8

       x = 3/8

Vậy x thuộc { 1/8 ; 3/8 }

c, 3 + | x+2 | = 2

          | x+2 | = 2 - 3 

          | x+2 | = -1 

<=> x + 2 = -1

       x + 2 = 1

<=> x = -1 - 2

       x = 1 -2 

<=> x = -3

       x = -1

Vậy x thuộc { -3 ; -1 }

\(a,\frac{7}{25}\cdot\frac{39}{-14}\cdot\frac{50}{78}=\frac{7}{25}\cdot-\frac{39}{14}\cdot\frac{25}{39}=-\frac{7}{14}=-\frac{1}{2}\)

\(b,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{6}{8}=\frac{3}{4}\)

Còn nữa:

7/19.8/11+3/11.7/19+-12/19

Bài 1:

a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)

\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)

\(=\frac{3}{4}-1-\frac{23}{5}\)

\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)

Bài 2:

a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)

\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)

\(2x+\frac{3}{4}=-1\)

\(2x=-1-\frac{3}{4}\)

\(2x=\frac{-7}{4}\)

x = -7/8

b, 3,2x - 2,7x + 8,5 = 6

x(3,2 - 2,7) = -2,5

0,5x = -2,5

x = -5

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

bn ơi cái này có r của hailey p336

ây da sai đề kìa bạn gì đó ơi

a: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)

b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)

a) \(\dfrac{6}{5\sqrt{8}}\)

\(=\dfrac{6}{5\cdot2\sqrt{2}}\)

\(=\dfrac{6}{10\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{10}\)

b) \(\dfrac{7}{5+2\sqrt{3}}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

\(=\dfrac{35-14\sqrt{3}}{13}\)

c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=3\sqrt{7}+3\sqrt{5}\)

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)