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a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
a, x-1:4/3+9/16=-5^(^laf mũ)2/4^2
x-1:4/3+9/16=25/16
x-1:4/3=25/16-9/16
x-1:4/3=1
x-1=1.4/3
x-1=4/3
x=4/3+1
x=7/3
kl:...................(hì tự viết nha mik lười viết)
b, 5/2x+2-31/2=1/6
5/2x+2=1/6+31/2
5/2x+2=16
5/2x=16-2
5/2x=14
x=14:5/2
x=28/5(vì 28 ko chia hết cho 5 nên mik viết như vậy)
*chúc bn học tốt nha* ^-^ *_*
\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)
\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)
\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)
\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)
\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)
\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-7}{14}\)
\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)
\(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)
\(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)
\(\frac{14}{9}\cdot x=-\frac{7}{18}\)
\(x=-\frac{7}{18}:\frac{14}{9}\)
\(x=-\frac{1}{4}\)
a) | x - 3 | - ( -3 ) = 4
| x - 3 | = 4 + ( -3 )
| x - 3 | = 1
=> x - 3 = -1 => x = ( -1 ) + 3 = -2
hoặc x - 3 = 1 => x = 1 + 3 = 4
vậy x = 1 hoặc x = 4
\(a)\left(3\frac{1}{2}-x\right).1\frac{1}{4}=\frac{15}{6}\)
\(\left(\frac{7}{2}-x\right).\frac{5}{4}=\frac{15}{6}\)
\(\frac{7}{2}-x=\frac{15}{6}:\frac{5}{4}\)
\(\frac{7}{2}-x=2\)
\(x=\frac{7}{2}-2\)
\(\Rightarrow x=\frac{3}{2}\)
\(\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\\frac{3}{4}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{3}{8}\end{cases}}\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{3}{8}\right\}\)
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