tìm x
\(\frac{9+x}{13-x}\)= \(\frac{5}{6}\)
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Tìm x, biết: \(\frac{9+x}{13-x}=\frac{5}{6}\)
\(\frac{9+x}{13-x}=\frac{5}{6}\)
=> (9 + x).6 = 5.(13 - x)
=> 54 + 6x = 65 - 5x
=> 6x + 5x = 65 - 54
=> 11x = 9
=> x = 9/11
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\(\frac{9+x}{13-x}=\frac{5}{6}\)
\(\left(9+x\right)\times6=\left(13-x\right)\times5\)
\(54+6\times x=65-5\times x\)
\(6\times x+5\times x=65-54\)
\(11\times x=11\)
\(x=1\)
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\(\frac{9+x}{13-x}=\frac{5}{6}\)
Tìm x
\(\frac{9+x}{13-x}=\frac{5}{6}\)
=> ( 9 + x ) .6 = ( 13 - x ) .5
=> 9.6 + 6x = 13.5 - 5x
=> 54 + 6x = 65 - 5x
=> 6x + 5x = 65 - 54
=> 11x = 11
=> x = 1
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Tìm x:
\(\frac{9+x}{13-x}=\frac{5}{6}\)
Tìm x \(\in\) Z
\(\frac{1}{-2}+\frac{1}{+}+\frac{-6}{9}\le x\le\left(\frac{1}{4}+\frac{-5}{13}\right)+\left(\frac{3}{6}+\frac{8}{-13}+\frac{9}{12}\right)\)
tìm x
a) \(x-\left[\frac{13}{18}x-\frac{24}{108}\right]=\left(\frac{-2}{3}\right)^2\)
b) \(3-7\frac{7}{12}< x< -\frac{5}{9}:\left(\frac{5}{9}-\frac{1}{6}\right)\)
\(\Leftrightarrow\)\(x-\left(\frac{13x}{18}-\frac{4}{18}\right)=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{18x}{18}-\frac{13x}{18}+\frac{4}{18}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{4}{9}-\frac{4}{18}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{2}{9}\)
\(\Leftrightarrow\)\(5x=\frac{18.2}{9}\)
\(\Leftrightarrow\)\(5x=4\)
\(\Leftrightarrow\)\(x=\frac{4}{5}\)
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Tìm x, biết: \(\frac{9+x}{13-x}=\frac{5}{6}\)
Có cách làm nữa nha các bạn
\(\frac{9+x}{13-x}=\frac{5}{6}\)
<=> (9+x).6 = (13 - x).5
<=> 54 + 6x = 65 - x.5
<=> 11x = 65 - 54
<=> x = 11 : 11
<=> x = 1
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\(\frac{9+x}{13-x}=\frac{5}{6}\)
=> 6(9 + x) = 5(13 - x)
=> 54 + 6x = 65 - 5x
=> 54 - 65 = - 5x + 6x
=> - 9 = - 1x
=> x = 9
vậy_
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Tìm x, biết:
\(x=\left(2-\frac{5}{3}+\frac{7}{6}-\frac{9}{10}+\frac{11}{15}-\frac{13}{21}+\frac{15}{28}-\frac{17}{36}+\frac{19}{45}\right)\times\frac{5}{3}\)
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tìm x biết:
a,\(7,5x:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)
b,\(\frac{\left(1,16-x\right)\cdot5,25}{\left(10\frac{5}{9}-7\frac{1}{4}\right)\cdot2\frac{2}{17}}\)=75%
Tìm x, biết:
a) \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
b)\(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
c) (x+2) - (x+3) >0
d)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
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