\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
đề : tìm x
Những câu hỏi liên quan
frac{12}{16}frac{-x}{4}frac{21}{y}frac{z}{80} (-0,6x-frac{1}{2}).frac{3}{4}-left(-1right)frac{1}{3}
frac{1}{3}x+frac{2}{5}left(x-1right)0
left(2x-3right).left(6-2xright)0
frac{-2}{3}-frac{1}{3}left(2x-5right)frac{3}{2}
2|frac{1}{2}x-frac{1}{3}|-frac{3}{2}frac{1}{4}
frac{3}{4}-2|2x-frac{2}{3}|2
Đọc tiếp
\(\frac{12}{16}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\) \((-0,6x-\frac{1}{2}).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\frac{-2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
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 \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
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Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
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a) \(\left(2x-3\right).\left(6-2x\right)=0\)
b) \(\frac{3}{4}-2.\left|2x-\frac{2}{3}\right|=2\)
c) \(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
Giúp mk vs nha
a) (2x - 3). (6 - 2x) = 0
==>2x - 3 = 0 hoặc 6 - 2x = 0
2x = 3 2x = 6
x = 3/2 x = 3
vậy x thuộc { 3/2 ; 3}
b) x không tồn tại
c) x = 35/54
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Bài1 tìm x
a)\(\left(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{2}{28.17}+\frac{2}{17.40}\right).x=\frac{3}{5}\)
b) \(6:\left(\frac{0,6x+8}{20}+9\right)=0,6\)
TÌM x
\(\left(\left(\frac{3}{4}\cdot x+5\right)-\left(\frac{2}{3}\cdot x-4\right)-\left(\frac{1}{6}\cdot x+1\right)\right)=\left(\frac{1}{3}\cdot x+4\right)-\left(\frac{1}{3}-3\right)\)
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
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2 tháng 8 2016 lúc 13:23
tìm x biết
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
tìm x biết :
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
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lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
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Tìm x : \(\frac{\left(x-3\right)^2}{2}-1\frac{1}{3}\left(x+2\right)^2-\frac{5}{4}\left(x-1\right)\left(x+1\right)=1\frac{1}{2}x\left(x-2\right)-x-4\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}\left(x^2+4x+4\right)-\dfrac{5}{4}\left(x^2-1\right)=\dfrac{3}{2}x\left(x-2\right)-x-4\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}x^2-\dfrac{16}{3}x-\dfrac{16}{3}-\dfrac{5}{4}x^2+\dfrac{5}{4}=\dfrac{3}{2}x^2-3x-x-4\)
\(\Leftrightarrow x^2\cdot\dfrac{-25}{12}-\dfrac{25}{3}x-\dfrac{103}{12}-\dfrac{3}{2}x^2+4x+4=0\)
\(\Leftrightarrow\dfrac{-43x^2}{12x}-\dfrac{13x}{3}-\dfrac{55}{12}=0\)
\(\Leftrightarrow43x^2+52x+55=0\)
\(\text{Δ}=52^2-4\cdot43\cdot55=-6756< 0\)
Do đó: Phương trình vô nghiệm
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1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tìm x
\(\frac{5}{2}-\left(\frac{3}{2}-2\frac{1}{3}+x\right)=\frac{8}{15}-\left(\frac{1}{4}-\frac{7}{10}\right)\)
\(1\frac{2}{3}-1\frac{3}{5}+x=\frac{2}{5}-\left|\frac{3}{4}-\frac{7}{8}\right|\)
\(2-\left(\frac{2}{3}-3\frac{1}{4}+x\right)=1\left|\frac{1}{6}-\frac{13}{12}\right|\)