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a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
d) \(x\left(x+1\right)-x-1=0\)
\(\Leftrightarrow x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Ta có: ( x - 2) x ( y + 3) = -13 = (-13) x 1 = (-1) x 13
* Nếu x - 2 = -13 => x = (-13) + 2 = -11
y + 3 = 1 => y = 1-3 = -2
* Nếu x-2 = -1 => x = (-1) + 2 = 1
y + 3 = 13 => y = 13 - 3 = 10
Vậy có 2 cặp x;y x;y(-11;-2)
x;y(1;10)
\(\frac{2}{3}x:\frac{1}{5}=\frac{4}{3}:25\%\)
\(\frac{2}{3}x:\frac{1}{5}=\frac{4}{3}:\frac{1}{4}=\frac{16}{3}\)
\(\frac{2}{3}x=\frac{16}{3}.\frac{1}{5}=\frac{16}{15}\)
\(x=\frac{16}{15}:\frac{2}{3}=\frac{8}{5}\)
Vậy x=8/5
\(\frac{x}{3}=\frac{5}{8}+\frac{1}{8}\)
\(\frac{x}{3}=\frac{6}{8}=\frac{3}{4}\)
\(\Rightarrow x\cdot4=3^2\)\(x=\frac{9}{4}\)hay\(x=2.25\)
Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^
duyệt đi
1) \(x^2+\frac{8}{9}=\frac{41}{36}\)\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)
2) \(\left(x-3\right)^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
3) \(\frac{3}{11}.\frac{22}{6}-\left(x-1\right)^2=\frac{7}{16}\)
\(\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{1}{4}\end{cases}}\)
4) \(1+\left(x+1\right)^3=\frac{37}{64}\)
\(\Leftrightarrow\left(x+1\right)^3=-\frac{27}{64}\)
\(\Rightarrow x+1=-\frac{3}{4}\)
\(\Leftrightarrow x=-\frac{7}{4}\)
5) \(\left(x-\frac{1}{2}\right)^2-\frac{9}{16}=1\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{25}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{4}\\x-\frac{1}{2}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-\frac{3}{4}\end{cases}}\)
6) Bn ghi rõ đề nha mk ko hiểu
6 ) \(\frac{3-x}{5-x}=\left(\frac{-3}{5}\right)^2\)
\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)
\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)