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a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
\(a,x\cdot\frac{1}{2}\cdot\frac{2}{3}=4\)
\(\Rightarrow x\cdot\frac{1}{3}=4\)
\(\Rightarrow x=12\)
\(b,-\frac{2}{7}\cdot\frac{5}{7}\cdot x=\frac{7}{21}\)
\(\Rightarrow-\frac{10}{49}x=\frac{7}{21}\)
\(\Rightarrow x=-\frac{49}{30}\)
k đi làm tiếp cho
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)
\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)
\(\Rightarrow2x-x=3-2\Rightarrow x=1\)
vay x=1
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)