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a, \(\frac{2}{5}+\frac{1}{4}\times x=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{3}{10}-\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{1}{4}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
Vậy \(x=\frac{-2}{5}\)
b, \(\frac{2}{3}+\frac{2}{3}\div x=\frac{4}{15}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{2}{3}\div\frac{-2}{5}\)
\(\Leftrightarrow\frac{-5}{3}\)
Vậy \(x=\frac{-5}{3}\)
c, \(2\times\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{2}\div2\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{4}\\\frac{2}{3}-x=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{11}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{12};\frac{11}{12}\right\}\)
d, \(3\times\left|\frac{5}{4}-x\right|-\frac{1}{8}=\frac{1}{4}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{1}{4}+\frac{1}{8}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{3}{8}\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{3}{8}\div3\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{1}{8}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-x=\frac{1}{8}\\\frac{5}{4}-x=\frac{-1}{8}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{9}{8}\\x=\frac{11}{8}\end{cases}}\)
Vậy \(x\in\left\{\frac{9}{8};\frac{11}{8}\right\}\)
a) 3/4+ 1/4:x = 2/5
1/4:x = 3/4-2/5
1/4:x= 7/20
x= 7/20:1/4
x= 7/5
b) chưa học
c) 15/8-1/8: (x/4 - 0,5) = 5/4
1/8: (x/4 -1/2)= 15/8-5/4
1/8:( x/4 -1/2) = 5/8
x/4 - 1/2 = 1/8:5/8
x/4 -1/2= 1/5
x/4= 1/5+1/2
x/4 = 7/7
x/4= 7/7× 4/4
x/4= 28/28
4/4=28/28
phần c ko chắc chắn
đúng k nhé
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)