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

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

=> \(\frac{2}{3}:x=\frac{1}{4}\)

=> \(x=\frac{2}{3}:\frac{1}{4}=\frac{8}{3}\)

b) \(5,4-3\left|x-\frac{21}{10}\right|=0\)

=> \(3\left|x-\frac{21}{10}\right|=\frac{27}{5}\)

=> \(\left|x-\frac{21}{10}\right|=\frac{27}{5}:3=\frac{9}{5}\)

=> \(\left|x-\frac{21}{10}\right|=\frac{9}{5}\)

Trường hợp 1 : \(x-\frac{21}{10}=\frac{9}{5}\)

=> \(x=\frac{9}{5}+\frac{21}{10}=\frac{39}{10}\)

Trường hợp 2 : \(x-\frac{21}{10}=-\frac{9}{5}\)

=> \(x=-\frac{9}{5}+\frac{21}{10}=\frac{3}{10}\)

Vậy : ...

c) \(10\sqrt{x-5}=25\)

=> \(\sqrt{x-5}=\frac{5}{2}\)

=> \(\left(x-5\right)^2=\frac{25}{4}\)

Trường hợp 1 :

\(x-5=\frac{25}{4}\)=> \(x=\frac{25}{4}+5=\frac{45}{4}\)

Trường hợp 2 :

\(x-5=-\frac{25}{4}\)=> \(x=-\frac{25}{4}+5=-\frac{5}{4}\)(loại) 

Vậy \(x=\frac{45}{4}\)

a,\(Đkxđ:x\ge3\)

Ta có:

\(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow|x-3|=3-x\)

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

\(TH1:x-3=x-3\Leftrightarrow0x=0\)

\(\Rightarrow\)\(x\in R\) và \(x\ge3\)

\(TH2:x-3=3-x\Leftrightarrow2x=6\Leftrightarrow x=3\)( ko thỏa mãn điều kiện)

vậy \(\left\{x\in R/x\ge3\right\}\)

b, \(Đkxđ:x\le\dfrac{5}{2}\)

Ta có:

\(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in R\\x=\dfrac{5}{2}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(\left\{x\in R/x\le\dfrac{5}{2}\right\}\)

Chúc bạn học tốt!

\(A=2-x\sqrt{\frac{x\left(x-2\right)}{\left(x-2\right)^2}+\frac{1}{\left(x-2\right)^2}}=2-x\sqrt{\frac{\left(x-1\right)^2}{\left(x-2\right)^2}}\)

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

\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)

Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)

Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)

\(C=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}+\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-5\right)^2}}=\frac{\sqrt{x}-5}{\sqrt{x}}+\left|\frac{\sqrt{x}-1}{\sqrt{x}-5}\right|\)

Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).