giải các bpt sau
$x^{2}$ + 5 $\geq$ 0
|x+5|= |1-$\frac{1}{3}$x |
|x+1| -1 -x =0
|x-3| + |x-1| =2
$x^{2}$ + |x-2| – 4x+ 4 +$y^{2}$ =0
giải các bpt sau $x^{2}$ + 5 $\geq$ 0 |x+5|= |1-$\frac{1}{3}$x | |x+1| -1 -x =0 |x-3| + |x-1| =2 $x^{2}$ + |x-2| – 4x+ 4 +$y^{2}$ =0
By Maria
Đáp án:
$\begin{array}{l}
a){x^2} + 5 \ge 0\left( {luon\,dung} \right)\\
b)\left| {x + 5} \right| = \left| {1 – \frac{1}{3}x} \right|\\
\Rightarrow \left[ \begin{array}{l}
x + 5 = 1 – \frac{1}{3}x\\
x + 5 = \frac{1}{3}x – 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\frac{4}{3}x = – 4\\
\frac{2}{3}x = – 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – 3\\
x = – 9
\end{array} \right.\\
c)\left| {x + 1} \right| – 1 – x = 0\\
\Rightarrow \left| {x + 1} \right| = x + 1\\
\Rightarrow x + 1 \ge 0\\
\Rightarrow x \ge – 1\\
d)\left| {x – 3} \right| + \left| {x – 1} \right| = 2\\
+ Khi:x \ge 3\\
\Rightarrow x – 3 + x – 1 = 2\\
\Rightarrow 2x = 6\\
\Rightarrow x = 3\left( {tmdk} \right)\\
+ Khi:1 \le x < 3\\
\Rightarrow 3 – x + x – 1 = 2\\
\Rightarrow 2 = 2\left( {luon\,dung} \right)\\
+ Khi:x < 1\\
\Rightarrow 3 – x + 1 – x = 2\\
\Rightarrow 2x = 2\\
\Rightarrow x = 1\left( {ktm} \right)\\
Vay\,1 \le x \le 3.\\
e){x^2} + \left| {x – 2} \right| – 4x + 4 + {y^2} = 0\\
\Rightarrow {\left( {x – 2} \right)^2} + \left| {x – 2} \right| + {y^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} \ge 0\\
\left| {x – 2} \right| \ge 0\\
{y^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} = 0\\
\left| {x – 2} \right| = 0\\
{y^2} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.
\end{array}$