Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x+\left(-1\right)-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=1-\frac{1}{3}-\frac{5}{6}\)
\(x.\left(-5-\frac{1}{2}-\frac{3}{2}\right)=\frac{-1}{6}\)
\(x.\left(-7\right)=\frac{-1}{6}\)
x=\(\frac{1}{42}\)
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
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: \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
a) (2x+3)-(5x-17)=9
2x+3-5x+17=9
-3x+20=9
-3x=9-20=-11
=>x=11/3
b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)
=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)
=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)
\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)
\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)
x=107/315:2/5=107/126
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=\frac{1}{5}-x\)
\(\left(3x-5x\right)-\left(\frac{3}{2}+3\right)=\frac{1}{5}-x\)
\(\left(-2\right).x-\frac{9}{2}=\frac{1}{5}-x\)
\(\left(-2\right).x+x.1=\frac{1}{5}+\frac{9}{2}\)
\(-1.x=\frac{47}{10}\)
\(x=\frac{47}{10}:\left(-1\right)\)
\(x=\frac{47}{10}.\frac{1}{-1}\)
\(x=\frac{47}{-10}\)
Vậy \(x=\frac{47}{-10}\)