1, giải các phương trình sau:
a, √x^2-6x+9 = x
b, x+ √ 4x^2-4x+1 =2
c, √ x+2 căn x+1 – √x-2 căn x +1 =2
d, √x+4 căn x+4 + √x-4 căn x +4 =4
1, giải các phương trình sau:
a, √x^2-6x+9 = x
b, x+ √ 4x^2-4x+1 =2
c, √ x+2 căn x+1 – √x-2 căn x +1 =2
d, √x+4 căn x+4 + √x-4 căn x +4 =4
\[\begin{array}{l}
a)\,\,\sqrt {{x^2} – 6x + 9} = x \Leftrightarrow \sqrt {{{\left( {x – 3} \right)}^2}} = x\\
\Leftrightarrow \left| {x – 3} \right| = x \Leftrightarrow \left[ \begin{array}{l}
x – 3 = x\\
x – 3 = – x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– 3 = 0\,\,\left( {vo\,\,ly} \right)\\
2x = 3
\end{array} \right. \Leftrightarrow x = \frac{3}{2}.\\
b)\,\,x + \sqrt {4{x^2} – 4x + 1} = 2 \Leftrightarrow \sqrt {{{\left( {2x – 1} \right)}^2}} = 2 – x\\
\Leftrightarrow \left| {2x – 1} \right| = 2 – x \Leftrightarrow \left[ \begin{array}{l}
2x – 1 = 2 – x\\
2x – 1 = x – 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 3\\
x = – 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = – 1
\end{array} \right..\\
c)\,\,\,\sqrt {x + 2\sqrt x + 1} – \sqrt {x – 2\sqrt x + 1} = 2\,\,\,\left( {DK:\,\,\,x \ge 0} \right)\\
\Leftrightarrow \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} – \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} = 1\\
\Leftrightarrow \left| {\sqrt x + 1} \right| – \left| {\sqrt x – 1} \right| = 1 \Leftrightarrow \sqrt x + 1 – \left| {\sqrt x – 1} \right| = 1\\
\Leftrightarrow \sqrt x = \left| {\sqrt x – 1} \right| \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \sqrt x – 1\\
\sqrt x = 1 – \sqrt x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0 = – 1\,\,\,\left( {VN} \right)\\
2\sqrt x = 1
\end{array} \right. \Leftrightarrow \sqrt x = \frac{1}{2} \Leftrightarrow x = \frac{1}{4}\,\,\,\left( {tm} \right).\\
d)\,\,\sqrt {x + 4\sqrt x + 4} + \sqrt {x – 4\sqrt x + 4} = 4\,\,\,\,\left( {DK:\,\,\,x \ge 0} \right)\\
\Leftrightarrow \sqrt {{{\left( {\sqrt x + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt x – 2} \right)}^2}} = 4\\
\Leftrightarrow \left| {\sqrt x + 2} \right| + \left| {\sqrt x – 2} \right| = 4\\
\Leftrightarrow \sqrt x + 2 + \left| {\sqrt x – 2} \right| = 4\\
\Leftrightarrow \left| {\sqrt x – 2} \right| = 4 – \sqrt x – 2\\
\Leftrightarrow \left| {\sqrt x – 2} \right| = 2 – \sqrt x \\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x – 2 = 2 – \sqrt x \\
\sqrt x – 2 = \sqrt x – 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2\sqrt x = 4\\
\,\,\forall x \ge 0
\end{array} \right.\\
\Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4.\\
Vay\,\,\,pt\,\,\,co\,\,nghiem\,\,voi\,\,moi\,\,x \ge 0.
\end{array}\]
Đáp án:
a. x=3/2
b.x= $ \pm $1
c.x=1
d.$0 \le x \le 4$
Giải thích các bước giải:
a. $\sqrt {{x^2} – 6x + 9} = x$
ĐKXĐ: $x \ge 0$
pttt: $\sqrt {{{\left( {x – 3} \right)}^2}} = x$
$\left| {x – 3} \right| = x$
x-3=x hoặc x-3= -x
0=3(vô lí) x=3/2 ( thỏa mãn ĐKXĐ )
Vậy x=3/2
b.
$\eqalign{ & x + \sqrt {4{x^2} – 4x + 1} = 2 \cr
& \sqrt {{{(2x – 1)}^2}} = 2 – x \cr} $
$\left| {2x – 1} \right| = 2 – x$ ( với $x \le 2$ )
khi 2x-1 $ \ge $0 $\eqalign{
& 2x – 1 = 2 – x \cr
& x = 1 \cr} $
khi 2x-1<0 $\eqalign{
& 2x - 1 = x - 2 \cr
& x = - 1 \cr} $
(thỏa mãn $x \le 2$)
Vậy x= $ \pm $1
c. ĐKXĐ x$ \ge $0
$\eqalign{
& \sqrt {x + 2\sqrt x + 1} - \sqrt {x - 2\sqrt x + 1} = 2 \cr
& \sqrt {{{(\sqrt x + 1)}^2}} - \sqrt {{{(\sqrt x - 1)}^2}} = 2 \cr
& \left| {\sqrt x + 1} \right| - \left| {\sqrt x - 1} \right| = 2 \cr
& \sqrt x + 1 - \left| {\sqrt x - 1} \right| = 2 \cr
& khi\sqrt x - 1 \ge 0 \cr
& \sqrt x + 1 - \sqrt x - 1 = 2(L) \cr
& khi\sqrt x - 1 < 0 \Leftrightarrow 0 \le x < 1 \cr & \sqrt x + 1 - 1 + \sqrt x = 2 \cr & 2\sqrt x = 2 \cr & x = 1 \cr} $ d.ĐKXĐ x$ \ge 0$ $\eqalign{ & \sqrt {x + 4\sqrt x + 4} + \sqrt {x - 4\sqrt x + 4} = 4 \cr & \sqrt {{{(\sqrt x + 2)}^2}} + \sqrt {{{(\sqrt x - 2)}^2}} = 4 \cr & \left| {\sqrt x + 2} \right| + \left| {\sqrt x - 2} \right| = 4 \cr & \sqrt x + 2 + \left| {\sqrt x - 2} \right| = 4 \cr & khi\sqrt x - 2 \ge 0 \Leftrightarrow x \ge 4 \cr & \sqrt x + 2 + \sqrt x - 2 = 4 \cr & 2\sqrt x = 4 \cr & x = 4(thoaman) \cr & khi\sqrt x - 2 < 0 \Leftrightarrow 0 \le x < 4 \cr & \sqrt x + 2 + 2 - \sqrt x = 4 \cr & 4 = 4 \cr} $ (luôn đúng) vậy $0 \le x \le 4$