- Tim x
\(50-2^3=\frac{x-5}{x}-23\)
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MT
10 tháng 6 2020
c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}.\frac{4}{3}\)
\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\).
d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)
\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).
H
1
25 tháng 3 2020
\(\frac{x-1}{4}=\frac{y+3}{4}=\frac{z-5}{4}=\frac{5z-25}{20}=\frac{3x-3}{12}=\frac{4y+12}{16}\)
\(=\frac{5z-3x-4y-25+3+20}{20-12-16}=\frac{50-2}{-8}=6\)
\(\Rightarrow x=6\cdot4+1=25\)
\(y=6\cdot4-3=21\)
\(z=6\cdot6+5=41\)
Vậy x=25;y=21;z=41
PM
1
CG
23 tháng 9 2019
-23/5.50/23≤x≤ -13/5:7/5
<=> -10≤x≤-13/7
Bạn tự chọn nốt số x nhé vì bạn ko cho điều kiện x thuộc gì
3 tháng 7 2017
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
\(50-2^3=\frac{x-5}{x}-23\)
=>\(50-8=1-\frac{5}{x}-23\)
=>\(42=-22+\frac{5}{x}\)
=>\(\frac{5}{x}=-22-42\)
=>\(\frac{5}{x}=-64\)
=>\(x=-\frac{5}{64}\)
\(50-2^3=\frac{x-5}{x}-23\)
\(\Rightarrow50-8=1-\frac{5}{x}-23\)
\(\Rightarrow1-\frac{5}{x}=42+23\)
\(\Rightarrow1-\frac{5}{x}=65\) \(\Rightarrow\frac{5}{x}=-64\Rightarrow x=-\frac{5}{64}\)