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f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x=\pm4\)(tmđk)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow10x^2+15x+5-30=0\)
\(\Leftrightarrow10x^2+15x-25=0\)
\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)
Vậy \(x=4\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)
\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)
\(\Leftrightarrow2x^2+3x+1=6\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)
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a) 169 . ( 3x - 9.17 ) + 24 : 3 = 30
169 . ( 3x - 153 ) + 8 = 30
169 . ( 3x - 153 ) = 30 - 8
169 . ( 3x - 153 ) = 22
3x - 153 = 22 : 169
3x - 153 = \(\frac{22}{169}\)
3x = \(\frac{22}{169}+153\)
3x = \(\frac{25879}{169}\)
x = \(\frac{25879}{169}:3\)
x = \(\frac{25879}{507}\)
Vậy \(x=\frac{25879}{507}\)
b) \(\left(\frac{4}{5}:\frac{6}{5}+\frac{1}{5}:\frac{1}{x}\right).30-26=54\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=54+26\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=80\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right)=80:30\)
\(\frac{2}{3}+\frac{1}{5}.x=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}-\frac{2}{3}\)
\(\frac{1}{5}.x=2\)
\(x=2:\frac{1}{5}\)
\(x=10\)
Vậy \(x=10\)
c) \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{18}\right):12\frac{1}{9}=0\)
\(\frac{1}{2}-\left(\frac{59}{9}+x-\frac{117}{18}\right):\frac{109}{9}=0\)
\(\frac{1}{2}.\left(\frac{59}{9}-\frac{117}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0:\frac{9}{109}\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0\)
\(\frac{1}{18}+x=0:\frac{1}{2}\)
\(\frac{1}{18}+x=0\)
\(x=0-\frac{1}{18}\)
\(x=\frac{-1}{18}\)
Vậy \(x=\frac{-1}{18}\)
d) 720 : [ 41 - ( 2x - 5 ) ] = 210
41 - 2x + 5 = 720 : 210
41 + 5 - 2x = \(\frac{24}{7}\)
46 - 2x = \(\frac{24}{7}\)
2x = \(46-\frac{24}{7}\)
2x = \(\frac{298}{7}\)
x = \(\frac{298}{7}:2\)
x = \(\frac{149}{7}\)
Vậy \(x=\frac{149}{7}\)
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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 \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
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a) ta có: \(\frac{x-3}{6}=\frac{3}{2x-6}\)
\(\Rightarrow\left(x-3\right).\left(2x-6\right)=6.3\)
\(\Rightarrow x.\left(2x-6\right)-3.\left(2x-6\right)=18\)
\(2x^2-6x-6x+18=18\)
\(2x^2-12x+18=18\)
\(2x^2-12x=0\)
\(2x.\left(x-6\right)=0\)
\(\Rightarrow2x=0\Rightarrow x=0\)
\(x-6=0\Rightarrow x=6\)
KL: x =0 hoặc x = 6
b) ta có: \(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\)
\(\Rightarrow\left(x+6\right).\left(2x-1\right)=\left(x+2\right).\left(2x+3\right)\)
\(\Rightarrow x.\left(2x-1\right)+6.\left(2x-1\right)=x.\left(2x+3\right)+2.\left(2x+3\right)\)
\(2x^2-x+12x-6=2x^2+3x+4x+6\)
\(2x^2+11x-6=2x^2+7x+6\)
\(\Rightarrow2x^2+11x-2x^2-7x=6+6\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
\(\frac{x-3}{6}=\frac{3}{2x-6}\Leftrightarrow\left(x-3\right).\left(2x-6\right)=18\Leftrightarrow2x^2-12x+18=18\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x.\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
\(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\Leftrightarrow\left(x+6\right).\left(2x-1\right)=\left(2x+3\right).\left(x+2\right)\)
\(\Leftrightarrow2x^2+11x-6=2x^2+7x+6\Leftrightarrow4x-12=0\Leftrightarrow x=3\)
\(\frac{2x+1}{6}=\frac{24}{2x+1}\)
Theo đề bài ta có:
\(\left(2x+1\right)\left(2x+1\right)=24.6=144\)
\(\Rightarrow\left(2x+1\right)^2=144\)
\(\Rightarrow2x+1=12\)
\(\Rightarrow2x=12-1\)
\(\Rightarrow2x=11\)
\(\Rightarrow x=\frac{11}{2}\)
\(\Rightarrow5,5\)
Vậy x = 5,5
Ta có: \(\frac{2x+1}{6}=\frac{24}{2x+1}\Rightarrow\left(2x+1\right)^2=24\times6=144\)
\(\Rightarrow2x+1=\sqrt{144}=12\)
\(\Rightarrow2x=12-1=11\Rightarrow x=\frac{11}{2}=5,5\)