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b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)
c) \(\left(3x+2\right)^3=-27\)
\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+2=-3\)
\(\Rightarrow3x=\left(-3\right)-2\)
\(\Rightarrow3x=-5\)
\(\Rightarrow x=\left(-5\right):3\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!

1.
a) \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
⇒ \(\left(x-2\right).\left(x+7\right)=\left(x+4\right).\left(x-1\right)\)
⇒ \(x^2+7x-2x-14=x^2-x+4x-4\)
⇒ \(x^2+7x-2x-14-x^2+x-4x+4=0\)
⇒ \(2x-10=0\)
⇒ \(2x=0+10\)
⇒ \(2x=10\)
⇒ \(x=10:2\)
⇒ \(x=5\)
Vậy \(x=5.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!


2x=16
Mà: 2.2.2.2=16
=>24=16
=> x=4
3x+1=9x
=>3x+1=32.x
=>x+1=2x
=>x=1
\(\dfrac{1}{2}x+1=\dfrac{9}{16}\\ \dfrac{1}{2}x=\dfrac{9}{16}-1\\ \dfrac{1}{2}x=\dfrac{-7}{16}\\ x=\dfrac{-7}{16}:\dfrac{1}{2}\\ x=\dfrac{-7}{16}\cdot2\\ x=\dfrac{-7}{8}\)
\(\left(\dfrac{1}{2}x+1\right)=\dfrac{9}{16}\)
\(\dfrac{1}{2}x=\dfrac{9}{16}-1\)
\(\dfrac{1}{2}x=\dfrac{-7}{16}\)
\(x=\left(\dfrac{-7}{16}\right)\div\dfrac{1}{2}\)
\(x=\dfrac{-7}{8}\)
Vậy \(x=\dfrac{-7}{8}\)