Giải phương trình sau: x^2 + (9x^2)/(x+3)^2=7 14/07/2021 Bởi Jade Giải phương trình sau: x^2 + (9x^2)/(x+3)^2=7
$x^{2}$ + $\frac{9x^2}{( x + 3)^2}$ = 7 (1) ĐK: x$\neq$ -3 (1) <=> $\frac{x^2.(x+3)^2 +9x^2 – 7.(x+3)^2}{(x+3)^2}$ = 0 <=> $x^2.(x+3)^2 +9x^2 – 7.(x+3)^2$ = 0 <=>$x^2.(x^2 +6x+9) +9x^2 – 7.(X^2 + 6x+9)$ =0 <=>$X^4 + 6x^3 + 9x^2 + 9x^2 -7x^2-42x – 63$=0 <=> $X^4 + 6x^3 + 11x^2 -42x-63 =0 $ <=> $(x^2 -x-3).(x^2 + 7x + 21) = 0$ <=> \(\left[ \begin{array}{l}x^2 -x-3 =0\\x^2 + 7x + 21 = 0 ( vô . nghiệm)\end{array} \right.\) <=> x = $\frac{1 + \sqrt[2]{13} }{2}$ hoặc x = $\frac{1 – \sqrt[2]{13} }{2}$ Bình luận
$\begin{array}{l} {x^2} + \dfrac{{9{x^2}}}{{{{\left( {x + 3} \right)}^2}}} = 7\\ \Leftrightarrow {\left( {x – \dfrac{{3x}}{{x + 3}}} \right)^2} + \dfrac{{2.3{x^2}}}{{x + 3}} = 7\\ \Leftrightarrow {\left( {\dfrac{{{x^2}}}{{x + 3}}} \right)^2} + \dfrac{{6{x^2}}}{{x + 3}} = 7\\ \Leftrightarrow {t^2} + 6t – 7 = 0\left( {t = \dfrac{{{x^2}}}{{x + 3}}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} t = – 7\\ t = 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} \dfrac{{{x^2}}}{{x + 3}} = – 7\\ \dfrac{{{x^2}}}{{x + 3}} = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + 7x + 21 = 0(PTVN)\\ {x^2} – x – 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt {13} }}{2}\\ x = \dfrac{{1 – \sqrt {13} }}{2} \end{array} \right. \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt {13} }}{2};\dfrac{{1 – \sqrt {13} }}{2}} \right\} \end{array}$ Bình luận
$x^{2}$ + $\frac{9x^2}{( x + 3)^2}$ = 7 (1)
ĐK: x$\neq$ -3
(1) <=> $\frac{x^2.(x+3)^2 +9x^2 – 7.(x+3)^2}{(x+3)^2}$ = 0
<=> $x^2.(x+3)^2 +9x^2 – 7.(x+3)^2$ = 0
<=>$x^2.(x^2 +6x+9) +9x^2 – 7.(X^2 + 6x+9)$ =0
<=>$X^4 + 6x^3 + 9x^2 + 9x^2 -7x^2-42x – 63$=0
<=> $X^4 + 6x^3 + 11x^2 -42x-63 =0 $
<=> $(x^2 -x-3).(x^2 + 7x + 21) = 0$
<=> \(\left[ \begin{array}{l}x^2 -x-3 =0\\x^2 + 7x + 21 = 0 ( vô . nghiệm)\end{array} \right.\)
<=> x = $\frac{1 + \sqrt[2]{13} }{2}$ hoặc x = $\frac{1 – \sqrt[2]{13} }{2}$
$\begin{array}{l} {x^2} + \dfrac{{9{x^2}}}{{{{\left( {x + 3} \right)}^2}}} = 7\\ \Leftrightarrow {\left( {x – \dfrac{{3x}}{{x + 3}}} \right)^2} + \dfrac{{2.3{x^2}}}{{x + 3}} = 7\\ \Leftrightarrow {\left( {\dfrac{{{x^2}}}{{x + 3}}} \right)^2} + \dfrac{{6{x^2}}}{{x + 3}} = 7\\ \Leftrightarrow {t^2} + 6t – 7 = 0\left( {t = \dfrac{{{x^2}}}{{x + 3}}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} t = – 7\\ t = 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} \dfrac{{{x^2}}}{{x + 3}} = – 7\\ \dfrac{{{x^2}}}{{x + 3}} = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + 7x + 21 = 0(PTVN)\\ {x^2} – x – 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt {13} }}{2}\\ x = \dfrac{{1 – \sqrt {13} }}{2} \end{array} \right. \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt {13} }}{2};\dfrac{{1 – \sqrt {13} }}{2}} \right\} \end{array}$