\(\frac{4x-3}{3}=\frac{3y+1}{7}=\frac{4x+3y-2}{5y}\)

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

\(=\frac{4x-3+3y+1-4x-3y+2}{10-5y}\)

\(=\frac{\left(4x-4x\right)+3y-3y-3+1+2}{10-5y}=0\)

\(\Rightarrow\hept{\begin{cases}4x-3=0\Leftrightarrow x=\frac{3}{4}\\3y+1=0\Leftrightarrow y=-\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{3}{4};y=-\frac{1}{3}\).

Câu trả lời đúng là :

x = 3/4

y = -1/3

Đáp số : ...

a) \(2.\left|5x-3\right|-2x=14\)

\(2\left|5x-3\right|=14+2x\)

\(\left|5x-3\right|=\frac{14+2x}{2}\)

\(\Rightarrow\orbr{\begin{cases}5x-3=\frac{-14-2x}{2}\\5x-3=\frac{14+2x}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\left(5x-3\right).2=-14-2x\\\left(5x-3\right).2=14+2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}10x-6+2x=-14\\10x-6-2x=14\end{cases}\Rightarrow\orbr{\begin{cases}12x=-14+6\\8x=14+6\end{cases}}}\Rightarrow\orbr{\begin{cases}12x=-8\\8x=20\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)

vậy \(\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)

Những câu sau tương tự nhé. 

làm giúp đi

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\) 

\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\) 

phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10

\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

giải chi tiết ra giúp mk nhé, cảm ơn nhiều