\(\dfrac{4^4-x}{3}=\dfrac{x-12}{5}\)

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21 tháng 10 2018

=> (\(^{4^4}\) -\(^x\) ).5=3.(\(^x\) -12)

=>(256-x).5=3.(x-12)

=> 1280-5x=3x-36

=> 1280 +36= 3x+5x

=> 1316= 8x

=> x= 1316:8

=> x =164,5

24 tháng 10 2018

THANK YOU!!!!

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)

=>-x-6-3=4

=>-x-9=4

=>-x=5

hay x=-5

b: =>(x+1)2=16

=>x+1=4 hoặc x+1=-4

=>x=3 hoặc x=-5

c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)

=>x-29=0

hay x=29

a: =>4x-6-9=5-3x-3

=>4x-15=-3x+2

=>7x=17

hay x=17/7

b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)

=>2/3x+21/3x=4/5+2+1/4=61/20

=>23/3x=61/20

=>3x=23:61/20=460/61

hay x=460/183

13 tháng 7 2017

Các câu dễ tự làm :v

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

8 tháng 6 2017

a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\dfrac{2}{5}+x=\dfrac{3}{12}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(x=\dfrac{-3}{20}\)

b)\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)

\(x=0:2\) \(x=0+\dfrac{1}{7}\)

\(x=0\) \(x=\dfrac{1}{7}\)

\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)

x= \(\dfrac{1.\left(-5\right)}{1.7}\)

\(x=\dfrac{-5}{7}\)

27 tháng 6 2017

a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)

\(x=\dfrac{8}{13}-\dfrac{3}{4}\)

\(x=-\dfrac{7}{52}\)

b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

c, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(2x-\dfrac{1}{7}=0\)

\(x-\dfrac{1}{7}=0:2\)

\(x-\dfrac{1}{7}=0\)

\(x=0-\dfrac{1}{7}\)

\(x=\dfrac{1}{7}\)

d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)

\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)

\(1:x=\dfrac{2}{5}\)

\(x=1:\dfrac{2}{5}\)

\(x=\dfrac{5}{2}\)

27 tháng 6 2017

a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)

b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)

c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)

\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)

\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)

vậy \(x=0;x=\dfrac{1}{7}\)

\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)

\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)

\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)

=>-6x=13/5

hay x=-13/30

22 tháng 4 2017

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22 tháng 4 2017

Ta có: \(\left|\dfrac{1}{4}+x\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}+x=\dfrac{5}{6}\\\dfrac{1}{4}+x=\dfrac{-5}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}-\dfrac{1}{4}\\x=\dfrac{-5}{6}-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{12}-\dfrac{3}{12}\\x=\dfrac{-10}{12}-\dfrac{3}{12}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)

Vậy \(x=\dfrac{7}{12}\) hoặc \(x=-\dfrac{13}{12}\)

23 tháng 9 2018

I , tìm x :

a, \(\left|x\right|=1,21\)

Ta có : \(\left|x\right|=\left|1,21\right|\rightarrow\left|x\right|=\pm1,21\)

b, \(\dfrac{11}{12}-\left(\dfrac{2}{5}-x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}-x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}-x=\dfrac{1}{4}\) => \(x=\dfrac{2}{5}-\dfrac{1}{4}\)

=> \(x=\dfrac{3}{20}\)

c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}\div x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}\div x=\dfrac{-7}{20}\) => \(x=\dfrac{1}{4}\div\dfrac{-7}{20}\)

=> \(x=\dfrac{-5}{7}\)

d,\(3^x=81\)

Ta có 81= \(3^4\)

Vì : \(3^x=3^4\Rightarrow x=4\)

e,\(\dfrac{1}{2}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)

\(\left|x\right|-\dfrac{5}{6}=\dfrac{8}{3}:\dfrac{1}{2}\)

=> \(\left|x\right|-\dfrac{5}{2}=\dfrac{16}{3}\) => \(\left|x\right|=\dfrac{16}{3}+\dfrac{5}{2}\)

=> \(\left|x\right|=\dfrac{47}{6}\)

\(\left|x\right|=\left|\dfrac{47}{6}\right|\Rightarrow x=\pm\dfrac{47}{6}\)

f, \(2^{x-3}=4\)

\(2^{x-3}=2^2\)

=> \(x-3=2\)

=> \(x=5\)

23 tháng 9 2018

a, Ta có \(\left|x\right|=1,21\)

\(\Rightarrow\left[{}\begin{matrix}x=1,21\\x=-1,21\end{matrix}\right.\)

Vậy \(x\in\left\{1,21;-1,21\right\}\)