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22 tháng 7 2018

\(3x\left(x-1\right)+5\left(2-x\right)=3x^2-7x+6\) \(6\)

<=> \(3x^2-3x+10-5x=3x^2-7x+6\)

<=> \(-x=-4\)

<=> \(x=4\)

\(\left(x+2\right)^2=\frac{1}{2}-\frac{1}{3}\)

<=> \(\left(x+2\right)^2=\frac{1}{6}\)

<=> \(\hept{\begin{cases}x+2=\sqrt{\frac{1}{6}}\\x+2=-\sqrt{\frac{1}{6}}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{6}}-2\\x=-\sqrt{\frac{1}{6}}-2\end{cases}}\)

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

5 tháng 6 2019

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

24 tháng 5 2016

a)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\left(\frac{x-1}{11}+3\right)+\left(\frac{x-1}{12}+2\right)=\left(\frac{x-1}{13}+3\right)+\left(\frac{x-1}{14}+2\right)\)

\(\left(\frac{x-1}{11}+\frac{x-1}{12}\right)+\left(3+2\right)=\left(\frac{x-1}{13}+\frac{x-1}{14}\right)+\left(3+2\right)\)

\(\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}+\frac{x-1}{14}=0\)

\(\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)\(\Rightarrow\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

24 tháng 5 2016

cái này là lớp 7 mà

phần a ra x=1

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

\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)

\(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0

=> x = -1

c) Xem lại đề

26 tháng 1 2017

Xin ỗi bạn nha! Đoạn x+3 sửa lại thành x+32 nha bạn !!!

4 tháng 11 2018

\(\frac{x+32}{11}=\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\Rightarrow\frac{x+32}{11}+\frac{x+23}{12}-\frac{x+38}{13}-\frac{x+27}{14}=0\)

\(\Rightarrow\left(\frac{x+32}{11}-\frac{x+38}{13}\right)+\left(\frac{x+23}{12}-\frac{x+27}{14}\right)=0\)

\(\Rightarrow\frac{2x-2}{11.13}+\frac{2x-2}{12.14}=0\)

\(\Rightarrow\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)

vì \(\left(\frac{1}{11.13}+\frac{1}{12.14}\right)\ne0\)

mà \(\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)

=> \(\frac{2x-2}{1}=0\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)