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a) \(\frac{-2}{3}\)- 3x = 0,75 + 5x
3x + 5x = \(\frac{-2}{3}\)- 0,75
8x = \(\frac{-17}{12}\)
x = \(\frac{-17}{12}\): 8
x =\(\frac{-17}{96}\)
Vậy x = \(\frac{-17}{96}\)
b) \(\frac{11}{12}\)- (\(\frac{2}{5}\)+ x ) = \(\frac{2}{3}\)
\(\frac{2}{5}\)+ x = \(\frac{11}{12}\)-\(\frac{2}{3}\)
\(\frac{2}{5}\)+ x = \(\frac{1}{4}\)
x = \(\frac{1}{4}\)- \(\frac{2}{5}\)
x = \(\frac{-3}{20}\)
Vậy x = \(\frac{-3}{20}\)
a) \(\frac{- 2}{3}\)- 3x = 0,75 + 5x
3x + 5x = \(\frac{- 2}{3}\)- 0,75
8x = \(\frac{- 17}{12}\)
x = \(\frac{- 17}{12}\): 8
x =\(\frac{- 17}{96}\)
Vậy x = \(\frac{- 17}{96}\)
b) \(\frac{11}{12}\)- (\(\frac{2}{5}\)+ x ) = \(\frac{2}{3}\)
\(\frac{2}{5}\)+ x = \(\frac{11}{12}\)-\(\frac{2}{3}\)
\(\frac{2}{5}\)+ x = \(\frac{1}{4}\)
x = \(\frac{1}{4}\)- \(\frac{2}{5}\)
x = \(\frac{- 3}{20}\)
Vậy x = \(\frac{- 3}{20}\)
sai thì thôi nhé hì

b) \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left|5x-3\right|=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy ....................
Bạn ơi !!! ý A tham khảo tại link này nè :
https://h.vn/hoi-dap/question/394208.html
~ Học tốt ~

\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)
\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)
\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)
Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:
\(E=5x.0+105=105\)

\(a,\left|5x+4\right|+7=26\\ \left|5x+4\right|=26+7\\ \left|5x+4\right|=33\\ \Rightarrow\left\{{}\begin{matrix}5x+4=33\\5x+4=-33\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x=29\\5x=-29\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\\x=-\dfrac{29}{5}\end{matrix}\right.\)
Các câu sau làm tương tự!

1.
b) \(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\)
Ta có:
\(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\ge\left(\left|x+8\right|+\left|-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left(\left|x+8-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left|-42\right|+\left|x+18\right|\)
\(\Rightarrow B=42+\left|x+18\right|\ge42\)
\(\Rightarrow MIN_B=42\) khi và chỉ khi:
\(\left\{{}\begin{matrix}x+8\ge0\\x+18=0\\x+50\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-8\\x=-18\\x\ge-50\end{matrix}\right.\Rightarrow x=-18.\)
Vậy \(MIN_B=42\) khi \(x=-18.\)
3.
b) \(\left|x-3\right|-\left|2x+1\right|=0\)
\(\Rightarrow\left|x-3\right|=\left|2x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=\left(-1\right)+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-1x=4\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4:\left(-1\right)\\x=2:3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-4;\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
- \(\frac{5}{17}\) x (\(\frac35x\) - 0,75) = 0
\(\frac35x\) - 0,75 = 0
\(\frac35\)\(x\) = 0,75
\(x=0,75:\frac35\)
\(x\) = \(\frac54\) \(\)
Vậy \(x=\frac54\)