Toán Lớp 10: Giải pt
$\sqrt[]{x^2+4x+4}$ – $\sqrt[]{x^2+2x-2}$ = 2
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 10: Giải pt
$\sqrt[]{x^2+4x+4}$ – $\sqrt[]{x^2+2x-2}$ = 2
Comments ( 2 )
\Leftrightarrow (x^2+4x+4) -2\sqrt{ (x^2+4x+4)(x^2+2x-2)} +(x^2+2x -2)=2^2 =4 \\
\Leftrightarrow 2x^2 +6x+2 -2\sqrt{ (x^2+4x+4)(x^2+2x-2)} =4 \\
\Leftrightarrow 2\sqrt{ (x^2+4x+4)(x^2+2x-2)} = 4 – (2x^2 +6x+2 ) =-2x^2-6x+2 \\
\Leftrightarrow (2\sqrt{ (x^2+4x+4)(x^2+2x-2)} )^2 = (-2x^2-6x+2 )^2 \\
\Leftrightarrow 4(x^4+6x^3+10x^2-8) =4x^4-24x^3+28x^2-24x+4 \\
\Leftrightarrow 4x^4+24x^3 +40x^2 -32 =4x^4+24x^3+28x^2-24x+4 \\
\Leftrightarrow (4x^4 -4x^4) +(24x^3 -24x^3 ) +(40x^2 -28x^2)+24x + (-32 -4) =0 \\
\Leftrightarrow 12x^2 +24x -36=0 \\
\Leftrightarrow 12(x^2 + 2x -3) =0 \\
\Leftrightarrow 12(x-1)(x+3) =0 \\
\Leftrightarrow \left[ \begin{array}{l}x-1 =0\\x+3 =0\end{array} \right.
\Leftrightarrow \left[ \begin{array}{l}x=1 ( \text{ N } )\\x=-3 ( \text{ L } )\end{array} \right.
\\
\text{Vậy } S = \{ 1 \} \end{array}$