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a: =>2x-1=4 hoặc 2x-1=-4
=>2x=5 hoặc 2x=-3
=>x=5/2 hoặc x=-3/2
d: =>x=|2|=2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)
\(a,|x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+3=\frac{1}{2}\\x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-7}{2}\end{cases}}\)
\(b,|2x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{1}{2}\\2x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=\frac{-7}{4}\end{cases}}\)
\(c,2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=\frac{-3}{2}\)
\(d,|2x+3|-1=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{2}\\2x+3=\frac{-3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{4}\\x=\frac{-9}{4}\end{cases}}\)
\(e,|2x+3|+5=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{-9}{2}\)(vô lí)
\(f,4-|2x+3|=1\Leftrightarrow|2x+3|=3\Leftrightarrow\orbr{\begin{cases}2x+3=3\\2x+3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
a, \(\left(x+1\right)^2=169\)
\(\left(x+1\right)^2=13^2\)
\(x+1=13\)
\(x=13-1\)
\(x=12\)
1.
a) \(\left(x+1\right)^2=169\)
⇒ \(x+1=\pm13\)
⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{12;-14\right\}.\)
b) \(\left(x+3\right)^3=-\frac{1}{27}\)
⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)
⇒ \(x+3=-\frac{1}{3}\)
⇒ \(x=\left(-\frac{1}{3}\right)-3\)
⇒ \(x=-\frac{10}{3}\)
Vậy \(x=-\frac{10}{3}.\)
c) \(\left(2x-4\right)^4=\frac{1}{625}\)
⇒ \(2x-4=\pm\frac{1}{5}\)
⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)
Còn câu d) bạn làm tương tự như mấy câu trên.
Chúc bạn học tốt!