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Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\\ \Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\\ \Leftrightarrow x=\frac{6}{9}\)
Vậy...
a) \(\left(x-3\right).\left(4-5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{3;\frac{4}{5}\right\}.\)
b) \(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0-\frac{1}{3}\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=-\frac{1}{3}.\)
Ta luôn có: \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x+\frac{3}{4}\right|>-\frac{1}{3}\)
\(\Rightarrow\left|x+\frac{3}{4}\right|\ne-\frac{1}{3}.\)
Vậy \(x\in\varnothing.\)
c) \(5^x.\left(5^3\right)^2=625\)
\(\Rightarrow5^x.5^6=5^4\)
\(\Rightarrow5^{x+6}=5^4\)
\(\Rightarrow x+6=4\)
\(\Rightarrow x=4-6\)
\(\Rightarrow x=-2\)
Vậy \(x=-2.\)
Chúc bạn học tốt!
a) \(5^x-\left(5^3\right)^2=625\)
\(\Leftrightarrow5^x-5^6=5^4\)
\(\Leftrightarrow x-6=4\)
\(\Leftrightarrow x=10\)
b) \(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)