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Có: \(4.\frac{-3}{10}\le x\le\frac{3}{11}.\frac{11}{30}\Rightarrow\frac{-6}{5}\le x\le\frac{1}{10}\)
\(\Rightarrow-\frac{12}{10}\le x\le\frac{1}{10}\) mà x là số nguyên \(\Rightarrow x=-1\)
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow\left|3x-\frac{1}{2}\right|=0\) \(\Rightarrow\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow3x-\frac{1}{2}=0\) \(\Rightarrow\frac{1}{2}y+\frac{3}{5}=0\)
\(3x=\frac{1}{2}\) \(\frac{1}{2}y=\frac{-3}{5}\)
\(x=\frac{1}{2}:3\) \(y=\left(\frac{-3}{5}\right):\frac{1}{2}\)
\(x=\frac{1}{6}\) \(y=\frac{-6}{5}\)
KL: x = 1/6; y = -6/5
b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|>0;\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
=> trường hợp |3/2x +1/9| + |1/5y -1/2| < 0 không thế xảy ra
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
rùi bn lm tương tự như phần a nhé!
\(a,\frac{7}{8}-\frac{1}{4}.\frac{5}{2}=\frac{x}{16}\)
\(\frac{7}{8}-\frac{5}{8}=\frac{x}{16}\)
\(\frac{2}{8}=\frac{x}{16}\)
\(\frac{4}{16}=\frac{x}{16}\)
=> X=4
k nha
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)
\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)
\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)
TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)
CHUC BAN HOC TOT :))
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}.\left(\left|-\frac{1}{3}\right|-\left|-\frac{1}{2}\right|-\left|-\frac{3}{-4}\right|\right)\)
\(\Leftrightarrow-\frac{13}{3}.\frac{1}{3}\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow-\frac{13}{9}\le x\le\frac{2}{4}.-\frac{11}{12}\)
\(\Leftrightarrow-\frac{13}{9}\le x\le-\frac{11}{24}\)
\(\Rightarrow x\in\left\{-1,0\right\}\) ( do \(x\in Z\) )
Vậy : \(x\in\left\{-1,0\right\}\)
\(-4\frac{1}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}\left(|\frac{-1}{3}|-|\frac{-1}{2}|-|\frac{-3}{-4}|\right)\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Rightarrow x\in\left[\frac{-13}{9};\frac{-11}{18}\right]\)
\(3\frac{2}{3}\left(\frac{1}{5}-\frac{1}{2}\right)\le x\le\frac{3}{11}\left(\frac{1}{5}+\frac{2}{3}-\frac{1}{2}\right)\)
\(\frac{11}{3}\left(\frac{1}{5}-\frac{1}{2}\right)\le x\le\frac{3}{11}\left(\frac{1}{5}+\frac{2}{3}-\frac{1}{2}\right)\)
\(-\frac{11}{10}\le x\le\frac{1}{10}\)
\(x=-1;0\)