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\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)
\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)
\(\Rightarrow x\in\){9:10;11;12;13;14}
\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)
\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)
Vậy \(x\in\left\{11;12;13\right\}\)
Câu 1,
x+y=-1/3 ; y+z=5/4 ; x+z= 4/3
=> 2(x+y+z)=9/4
=> x+y+z=9/8
Ta lại có: x+y=-1/3
=> z=9/8 -(-1/3)=35/24
Ta lại có: z+y=5/4
=> y=-5/24
=> x=.....
Câu 2:
\(-4\le x\le-\frac{11}{18}\)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
tìm x,y,z thuộc Q biết
\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
Xét đẳng thức , ta thấy :
\(\left|x+\frac{3}{4}\right|\ge0\)
\(\left|y-\frac{1}{5}\right|\ge0\)
\(\left|x+y+z\right|\ge0\)
=> \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\)
Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\) (đề bài)
=> \(\hept{\begin{cases}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=-\left(-\frac{3}{4}+\frac{1}{5}\right)=\frac{11}{20}\end{cases}}\)