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PH
4
XO
17 tháng 4 2020
Ta có : \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
=> (x - 1)(x + 3) = (x - 2)(x + 2)
=> x2 + 2x - 3 = x2 - 4
=> 2x = - 1
=> x = -0,5
17 tháng 4 2020
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\left(x\ne-2;x\ne-3\right)\)
<=>(x-1)(x+3)=(x-2)(x+2)
<=>x2+2x-3=x2-4
<=> x2+2x-3-x2+4=0
<=> 2x+1=0
<=> x=\(\frac{-1}{2}\)(tm)
5 tháng 6 2022
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
VM
17 tháng 10 2019
a) \(x^2-2=0\)
\(\Rightarrow x^2-\left(\sqrt{2}\right)^2=0\)
\(\Rightarrow\left(x-\sqrt{2}\right).\left(x+\sqrt{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\x+\sqrt{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+\sqrt{2}\\x=0-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}.\)
b) \(x^2+\frac{7}{4}=\frac{23}{4}\)
\(\Rightarrow x^2=\frac{23}{4}-\frac{7}{4}\)
\(\Rightarrow x^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}.\)
c) \(\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2=0^2\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=0+1\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
g) \(\sqrt{x}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0.\)
h) \(\sqrt{x}=4\)
\(\Rightarrow\sqrt{x}=\left(\sqrt{4}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{16}\)
\(\Rightarrow x=16\)
Vậy \(x=16.\)
i) \(\sqrt{x}-\frac{1}{7}=0\)
\(\Rightarrow\sqrt{x}=0+\frac{1}{7}\)
\(\Rightarrow\sqrt{x}=\frac{1}{7}\)
\(\Rightarrow\sqrt{x}=\left(\sqrt{\frac{1}{7}}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\frac{1}{49}}\)
\(\Rightarrow x=\frac{1}{49}\)
Vậy \(x=\frac{1}{49}.\)
Chúc bạn học tốt!
27 tháng 8 2017
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=1^2\)
\(x-2=1\)
\(x=1+2\)
\(x=3\)
c) \(\left(2x-1\right)^3=\left(-8\right)\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=\left(-2\right)\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}\)
27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^2=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)
\(\left(x+1\right)^2=\left(x+1\right)^0\)
\(\Rightarrow\left(x+1\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)