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17 tháng 7 2019
a) \(\frac{4}{x+5}=\frac{3}{2x-1}\)
=> 4(2x - 1) = 3(x + 5)
=> 8x - 4 = 3x + 15
=> 8x - 3x = 15 + 4
=> 5x = 19
=> x = 19/5
b) \(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)
=> \(\left(\frac{x+11}{19}-1\right)+\left(\frac{x+12}{20}-1\right)+\left(\frac{x+13}{21}-1\right)=0\)
=> \(\frac{x-8}{19}+\frac{x-8}{20}+\frac{x-8}{21}=0\)
=> \(\left(x-8\right)\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
=> x - 8 = 0
=> x = 8
c) \(\left(2x-1\right)^2=\left(2x-1\right)^3\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^3=0\)
=> \(\left(2x-1\right)^2.\left[1-\left(2x-1\right)\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\1-2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
17 tháng 7 2019
a) 4/x + 3 = 3/2x - 1
<=> 4.(2x - 1) = (x + 3).3
<=> 8x - 4 = 3x + 9
<=> 8x = 3x + 9 + 4
<=> 8x = 3x + 13
<=> 8x - 3x = 13
<=> 5x = 13
<=> x = 13/5
=> x = 13/5
c) (2x - 1)2 = (2x - 1)3
<=> 4x2 - 4x + 1 = 8x3 - 12x2 + 6x - 1
<=> 8x3 - 12x2 + 6x - 1 = 4x2 - 4x + 1
<=> 8x3 - 12x2 + 6x - 1 - 1 = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x - 4x = 4x2
<=> 8x3 - 12x2 + 10x - 2 = 4x2
<=> 8x3 - 12x2 + 10x - 2 - 4x2 = 0
<=> 8x2 - 16x2 + 10x - 2 = 0
<=> 2(x - 1)(2x - 1)2 = 0
<=> x - 1 = 0 hoặc 2x - 1 = 0
x = 0 + 1 2x = 0 + 1
x = 1 2x = 1
x = 1/2
=> x = 1 hoặc x = 1/2
6 tháng 7 2019
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
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 \(x\in\left\{\frac{1}{4}\right\}\)
17 tháng 7 2016
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
=> 7/2 : |2x-1| = 21/22
=> |2x-1| = 7/2 : 21/22 = 7/2 . 22/21 = 11/3
=> 2x-1=-11/3 hoặc 2x-1=11/3
=> x=-4/3 hoặc 7/3
Tk mk nha