\(\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}+-\dfrac{3}{5}:x=0\end{matrix}\right.\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\-\dfrac{3}{5}:x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{36}{48}\\x=-\dfrac{9}{5}\end{matrix}\right.\)

lâu rồi ko lm ko bt đúng ko

Ta có: \(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{-3}{5}:x+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{-3}{5}:x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\dfrac{-3}{5}:x=\dfrac{-1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-3}{5}:\dfrac{-1}{3}=\dfrac{-3}{5}\cdot\dfrac{3}{-1}=\dfrac{9}{5}\end{matrix}\right.\)

a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)

\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)

\(\dfrac{19}{20}x=1\)

\(x=\dfrac{20}{19}\)

Vậy \(x=\dfrac{20}{19}\)

b) \(\left(x^2-9\right)\left(3-5x\right)=0\)

TH1:

\(x^2-9=0\)

\(x^2=9\)

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

=>\(x\in\left\{3;-3\right\}\)

TH2:

\(3-5x=0\)

\(5x=3\)

\(x=\dfrac{3}{5}\)

Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)

bạn viết lại câu b được không?

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{3}{5}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}-\dfrac{3}{5}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\dfrac{3}{5}:x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{9}{5}\end{matrix}\right.\)

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{3}{5}:x\right)=0\)

\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{5}{3}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}-\dfrac{5}{3}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\Rightarrow x=\dfrac{3}{4}\\\dfrac{5}{3}x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)

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

\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)

\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)

c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)

\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)

hay \(x=-\dfrac{26}{21}\)

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

\(=\dfrac{-1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)

\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)

\(\Leftrightarrow6x+6+12x-8=x-7\)

\(\Leftrightarrow6x+12x-x=-7-6+8\)

\(\Leftrightarrow17x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{17}\)

Vậy .........................

b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)

\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)

\(\Leftrightarrow2x^2-x^2+x+15-21=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2-2x+3x-6=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy .........................

P/s: các câu còn lại tương tự, bn tự giải nha