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\(2.\)
\(a,\frac{1}{3}< \frac{x}{4}< \frac{2}{3}\)
\(\Rightarrow\frac{1\times4}{3\times4}< \frac{x\times3}{4\times3}< \frac{2\times4}{3\times4}\)
\(\Rightarrow\frac{4}{12}< \frac{x\times3}{12}< \frac{8}{12}\)
\(\Rightarrow4< x\times3< 8\)
\(\Rightarrow x\times3\in\left\{5;6;7\right\}\)
Ta có bảng giá trị :
| \(x\times3\) | \(5\) | \(6\) | \(7\) |
| \(x\) | \(\varnothing\) | \(2\) | \(\varnothing\) |
| \(NX\) | Loại | TM | Loại |
Vậy \(x=2\)
\(b,\frac{3}{5}< \frac{-x}{3}< \frac{4}{5}\)
\(\Rightarrow\frac{3\times3}{5\times3}< \frac{-x\times5}{3\times5}< \frac{4\times3}{5\times3}\)
\(\Rightarrow\frac{9}{15}< \frac{-x\times5}{15}< \frac{12}{15}\)
\(\Rightarrow9< -x\times5< 12\)
\(\Rightarrow-x\times5\in\left\{10;11\right\}\)
Ta có bảng giá trị :
| \(-x\times5\) | \(10\) | \(11\) |
| \(-x\) | \(2\) | \(\varnothing\) |
| \(x\) | \(-2\) | \(\varnothing\) |
| \(NX\) | TM | Loại |
Vậy \(x=-2\)
\(A=\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6+\frac{7}{4}+\frac{3}{2}\right)\)
\(A=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6-\frac{7}{4}-\frac{3}{2}\)
\(A=\left(3-5-6\right)-\left(\frac{1}{4}+\frac{7}{4}+\frac{3}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{6}{5}\)
\(A=-8-\left(2+\frac{3}{2}\right)+1+\frac{6}{5}\)
\(A=-8-2-\frac{3}{2}+1+\frac{6}{5}\)
\(A=-9-\frac{3}{2}+\frac{6}{5}\)
\(A=\frac{-93}{10}\)
Mk lm đc 1 cách thui
Ủng hộ mk nha ^_-
a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
=>(1/2x+9/4).(-2/3)=17/6
=>-2/3.1/2x+-2/3.9/4=17/6
=>-1/3x+3/2=17/6
=>-1/3x=17/6-3/2
=>-1/3x=4/3
=>x=(4/3)/-1/3
=>x=-4
1) Ta có: \(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|-\frac{3}{2}=\frac{1}{4}\)
⇔\(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
⇔\(\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{7}{4}:2=\frac{7}{4}\cdot\frac{1}{2}=\frac{7}{8}\)
⇔\(\left[{}\begin{matrix}\frac{1}{2}x-\frac{3}{8}=\frac{7}{8}\\\frac{1}{2}x-\frac{3}{8}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{10}{8}\\\frac{1}{2}x=\frac{-4}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{10}{8}:\frac{1}{2}=\frac{10}{8}\cdot2=\frac{20}{8}=\frac{5}{2}\\x=\frac{-4}{8}:\frac{1}{2}=-\frac{4}{8}\cdot2=-\frac{8}{8}=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{2};-1\right\}\)
2) Ta có: \(-5\cdot\left(x+\frac{1}{5}\right)-\frac{1}{2}\cdot\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
⇔\(-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{3}{2}x+\frac{5}{6}=0\)
\(\Leftrightarrow-7x+\frac{1}{6}=0\)
\(\Leftrightarrow-7x=-\frac{1}{6}\)
hay \(x=\frac{1}{42}\)
Vậy: \(x=\frac{1}{42}\)
3) Ta có: \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-\frac{3}{2}-5x-3+x-\frac{1}{5}=0\)
\(\Leftrightarrow-x-\frac{47}{10}=0\)
⇔\(-x=\frac{47}{10}\)
hay \(x=\frac{-47}{10}\)
Vậy: \(x=\frac{-47}{10}\)
4) Ta có: \(\frac{3}{4}-2\left|2x-0,125\right|=2\)
\(\Leftrightarrow2\left|2x-\frac{1}{8}\right|=\frac{3}{4}-2=-\frac{5}{4}\)
⇔\(\left|2x-\frac{1}{8}\right|=-\frac{5}{8}\)(vô lý)
Vậy: x∈∅
5) Ta có: \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
⇔\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\\\frac{1}{2}x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{24}:\frac{1}{2}=\frac{29}{24}\cdot2=\frac{29}{12}\\x=-\frac{13}{24}:\frac{1}{2}=-\frac{13}{24}\cdot2=-\frac{13}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}\)
\(=\frac{8}{8}=1\)
\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8
=\(\frac{327}{722}\)+5/8
=\(\frac{3113}{2888}\)
a, \(\left(x-\frac{2}{3}\right)^2=\frac{5}{6}\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4+\sqrt{30}}{6}\\x=\frac{4-\sqrt{30}}{6}\end{cases}}}\)nghiệm xấu thế ?
b, \(\left(\frac{3}{4}-x\right)^3=\left(-8\right)\Leftrightarrow\frac{3}{4}-x=-2\Leftrightarrow x=\frac{11}{4}\)