\(\left(\frac{3}{5^{ }}\right)^{2x^{ }}=\frac{9}{25}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)
\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)
\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)
\(\Leftrightarrow\text{ }x=\frac{53}{45}\)
\(\text{Vậy }x=\frac{53}{45}\)
\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)
\(\text{TH 1: Dấu chia}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)
\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)
\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)
\(\text{Vậy }\text{}x=\frac{-1}{5}\)
\(\text{TH 2:Dấu cộng}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{-3}{25}\)
\(\Leftrightarrow x=\frac{-3}{25}:2\)
\(\Leftrightarrow x=\frac{-3}{50}\)
\(\text{Vậy }x=\frac{-3}{50}\)
\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)
\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{5}{4}\)
\(\text{Vậy }x=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)
b)
\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_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_
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\left(\frac{3}{5}\right)^{2x}=\frac{9}{25}\)
<=> \(\left(\frac{3}{5}\right)^{2x}=\left(\frac{3}{5}\right)^2\)
<=> 2x = 2
<=> x = 1
\(\left(\frac{3}{5}\right)^{2x}=\frac{9}{25}\)
=> \(\left(\frac{3}{5}\right)^{2x}=\frac{3}{5}^2\)
=> 2x = 2
=> x = 2:2
=> x = 1