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\(12.\left(\frac{3}{4}-5x\right)=12.\frac{3}{4}-12.5x=9-12.5x=9-60x\)
\(9-60x=\frac{4}{5}-3x\), Rồi bạn làm bước cuối, chuyển vế
\(2\frac{5}{7}+\frac{3}{14}-\frac{6}{7}+0,5+\frac{9}{14}=\frac{19}{7}+\frac{3}{14}-\frac{6}{7}+0,5+\frac{9}{14}\)
\(=\left(\frac{19}{7}-\frac{6}{7}\right)+\left(\frac{3}{14}+\frac{9}{14}\right)+0,5=1+\frac{6}{7}+0,5=\frac{33}{14}\)

b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....

Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!

1
2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x
=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)
=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)
=> \(x=\frac{1}{10}\)
2 .
\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
=>\(11x=\frac{1}{6}\)
=>x=\(\frac{1}{66}\)
3.
\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
=>\(-2x=\frac{-2}{3}\)
=>\(\frac{1}{3}\)
4. câu 4 ko hiểu bạn ơi

\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Leftrightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\frac{4}{5}-\frac{3}{2}\)
\(\Leftrightarrow-7x=-\frac{7}{10}\)
\(\Leftrightarrow x=\frac{1}{10}\)

\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)

Bài làm :
a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)
b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)
c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)
d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)
e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)
\(2\times\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(-10x+3x=\frac{4}{5}-\frac{3}{2}\)
\(-7x=-\frac{7}{10}\)
\(x=\frac{1}{10}\)
\(2\times\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Rightarrow\frac{4}{5}-3x+10x=\frac{3}{2}\)
\(\Rightarrow\frac{4}{5}-7x=\frac{3}{2}\)
\(\Rightarrow7x=\frac{4}{5}-\frac{3}{2}\)
\(\Rightarrow7x=\frac{-7}{10}\)
\(\Rightarrow x=\frac{\frac{-7}{10}}{7}\)
\(\Rightarrow x=\frac{-1}{10}\)