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d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
a, Ta có:
\(2.x-\frac{1}{2}=\frac{5}{4}-\frac{1}{2}.x\)
\(\Leftrightarrow2.x+\frac{1}{2}.x=\frac{5}{4}+\frac{1}{2}\)
\(\Leftrightarrow\frac{5}{2}.x=\frac{7}{4}\)
\(\Leftrightarrow x=\frac{7}{10}\). Vậy: \(x=\frac{7}{10}\)
b, Ta có:
\(\frac{3}{4}-\frac{5}{2}.x=2-3.x\)
\(\Leftrightarrow2-\frac{3}{4}=3.x-\frac{5}{2}.x\)
\(\Leftrightarrow\frac{5}{4}=\frac{1}{2}.x\)
\(\Leftrightarrow x=\frac{5}{2}\). Vậy: \(x=\frac{5}{2}\)
c, Ta có:
\(\frac{3}{2}.\left(x-\frac{1}{3}\right)-2.\left(x-\frac{1}{2}\right)=3\)
\(\Leftrightarrow\frac{3}{2}.x-\frac{3}{2}.\frac{1}{3}-\left(2.x-2.\frac{1}{2}\right)=3\)
\(\Leftrightarrow\frac{3}{2}.x-\frac{1}{2}-2.x+1=3\)
\(\Leftrightarrow-\frac{1}{2}.x+\frac{1}{2}=3\)
\(\Leftrightarrow-\frac{1}{2}.x=\frac{5}{2}\)
\(\Leftrightarrow x=-5\). Vậy: \(x=-5\)
d, Ta có:
\(\frac{4}{3}.\left(x-\frac{3}{2}\right)-3.\left(\frac{1}{3}.x+1\right)=6\)
\(\Leftrightarrow\frac{4}{3}.x-\frac{4}{3}.\frac{3}{2}-3.\frac{1}{3}.x-3=6\)
\(\Leftrightarrow\frac{4}{3}.x-2-x-3=6\)
\(\Leftrightarrow\frac{1}{3}.x-5=6\)
\(\Leftrightarrow\frac{1}{3}.x=11\)
\(\Leftrightarrow x=33\). Vậy: \(x=33\)
Chúc bạn học tốt!Tick cho mình nhé!
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
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: \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Việt Nam đất nước anh hùng.....^^
Trung Quốc là nước nửa khùng nửa điên.
Việt Nam đang sống bình yên.
Trung Quốc đừng có làm phiền Việt Nam.
Trung Quốc đông dân toàn cỏ rác.
Việt Nam lác đác toàn siêu nhân.
Việt Nam cưỡi rồng bay trong gió.
Trung Quốc cưỡi chó sủa:"gâu" "gâu".
Thái Lan hỏi nó đi đâu.
Nó cười, nó bảo:" đi hầu Việt Nam
\(\frac{1}{2}x+\frac{2}{3}\left(x+2\right)=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{2}{3}x+\frac{4}{3}=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1}{2}+\frac{2}{3}\right)x+\frac{4}{3}=\frac{1}{3}\)
\(\Leftrightarrow\frac{7}{6}x+\frac{4}{3}=\frac{1}{3}\)
\(\Leftrightarrow\frac{7}{6}x=\frac{1}{3}-\frac{4}{3}\)
\(\Leftrightarrow\frac{7}{6}x=-1\)
\(\Leftrightarrow x=\left(-1\right):\frac{7}{6}\)
\(\Leftrightarrow x=\frac{-6}{7}\)
Vậy x = \(\frac{-6}{7}\)
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