\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
đề : tìm x
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30 tháng 4 2019
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
30 tháng 4 2019
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
31 tháng 3 2017
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
2 tháng 8 2016
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
12 tháng 8 2016
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
3 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}\left(x^2+4x+4\right)-\dfrac{5}{4}\left(x^2-1\right)=\dfrac{3}{2}x\left(x-2\right)-x-4\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}x^2-\dfrac{16}{3}x-\dfrac{16}{3}-\dfrac{5}{4}x^2+\dfrac{5}{4}=\dfrac{3}{2}x^2-3x-x-4\)
\(\Leftrightarrow x^2\cdot\dfrac{-25}{12}-\dfrac{25}{3}x-\dfrac{103}{12}-\dfrac{3}{2}x^2+4x+4=0\)
\(\Leftrightarrow\dfrac{-43x^2}{12x}-\dfrac{13x}{3}-\dfrac{55}{12}=0\)
\(\Leftrightarrow43x^2+52x+55=0\)
\(\text{Δ}=52^2-4\cdot43\cdot55=-6756< 0\)
Do đó: Phương trình vô nghiệm
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}+1=\frac{1}{3}\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{1}{3}-1\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{-2}{3}\)
\(-0,6x-\frac{1}{2}=-\frac{2}{3}:\frac{3}{4}\)
\(-0,6x=-\frac{8}{9}\)
\(x=\frac{-8}{9}:\left(-0,6\right)\)
\(x=\frac{40}{27}\)
@SKT_NTT cảm ơn bn