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a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{1}{2}\)
c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)
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Lời giải:
Ta có: $(x-2)^2\geq 0$ với mọi $x$
$|y^2-4|\geq 0$ theo tính chất trị tuyệt đối
Do đó $(x-2)^2+|y^2-4|\geq 0$. Để tổng $(x-2)^2+|y^2-4|=0$ thì:
$(x-2)^2=|y^2-4|=0$
$\Rightarrow x=2; y=\pm 2$
Ta có (x - 2)^2 + |y^2 - 4| = 0 (1)
Mà \(\left(x-2\right)^2\ge0,\left|y^2-4\right|\ge0\) với mọi x,y nên (1) xảy ra <=>
(x - 2)^2 = |y^2 - 4| = 0 <=> x - 2 = y^2 - 4 = 0 <=> x = 2 và y = 2,-2
Vậy...
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/5x-4/=/x+2/
\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)
vậy x=3/2 hoặc x=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
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\(x\left(x-\frac{1}{3}\right)< 0\)
Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau
Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$4|x+50\text{%}|-2=0$
$4|x+50\text{%}|=2$
$|x+50\text{%}|=\frac{1}{2}$
$\Rightarrow x+50\text{%}=\pm \frac{1}{2}$
$\Rightarrow x+\frac{1}{2}=\pm \frac{1}{2}$
$\Rightarrow x=0$ hoặc $x=-1$