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TQ
0
28 tháng 7 2019
\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
\(\Rightarrow\)\(\frac{x+32}{11}-3+\frac{x+23}{12}-2=\frac{x+38}{13}-3+\frac{x+27}{14}-2\)
\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)
\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}-\frac{x-1}{14}=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Rightarrow x-1=0\)(Vì \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\))
\(\Rightarrow x=1\)
Vậy:x=1
NL
0
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
25 tháng 1 2017
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
TN
22 tháng 7 2018
\(3x\left(x-1\right)+5\left(2-x\right)=3x^2-7x+6\) \(6\)
<=> \(3x^2-3x+10-5x=3x^2-7x+6\)
<=> \(-x=-4\)
<=> \(x=4\)
\(\left(x+2\right)^2=\frac{1}{2}-\frac{1}{3}\)
<=> \(\left(x+2\right)^2=\frac{1}{6}\)
<=> \(\hept{\begin{cases}x+2=\sqrt{\frac{1}{6}}\\x+2=-\sqrt{\frac{1}{6}}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{6}}-2\\x=-\sqrt{\frac{1}{6}}-2\end{cases}}\)
\(\frac{x+32}{11}=\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
\(\Rightarrow\frac{x+32}{11}+\frac{x+23}{12}-\frac{x+38}{13}-\frac{x+27}{14}=0\)
\(\Rightarrow\left(\frac{x+32}{11}-\frac{x+38}{13}\right)+\left(\frac{x+23}{12}-\frac{x+27}{14}\right)=0\)
\(\Rightarrow\frac{2x-2}{11.13}+\frac{2x-2}{12.14}=0\)
\(\Rightarrow\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)
vì \(\left(\frac{1}{11.13}+\frac{1}{12.14}\right)\ne0\)
mà \(\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)
=> \(\frac{2x-2}{1}=0\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)