Giải phương trình: (x+3)$\sqrt[2]{x+4}$ + (x+9)$\sqrt[2]{x+11}$ = $x^{2}$ + 9x +10 06/08/2021 Bởi Serenity Giải phương trình: (x+3)$\sqrt[2]{x+4}$ + (x+9)$\sqrt[2]{x+11}$ = $x^{2}$ + 9x +10
Đáp án: \[x = 5\] Giải thích các bước giải: ĐKXĐ: \(x \ge – 4\) Ta có: \(\begin{array}{l}\left( {x + 3} \right)\sqrt {x + 4} + \left( {x + 9} \right)\sqrt {x + 11} = {x^2} + 9x + 10\\ \Leftrightarrow \left( {x + 3} \right).\left( {\sqrt {x + 4} – 3} \right) + \left( {x + 9} \right)\left( {\sqrt {x + 11} – 4} \right) = {x^2} + 9x + 10 – 3\left( {x + 3} \right) – 4\left( {x + 9} \right)\\ \Leftrightarrow \left( {x + 3} \right).\frac{{x + 4 – {3^2}}}{{\sqrt {x + 4} + 3}} + \left( {x + 9} \right).\frac{{x + 11 – {4^2}}}{{\sqrt {x + 11} + 4}} = {x^2} + 2x – 35\\ \Leftrightarrow \frac{{\left( {x + 3} \right)\left( {x – 5} \right)}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 9} \right)\left( {x – 5} \right)}}{{\sqrt {x + 11} + 4}} = \left( {x – 5} \right)\left( {x + 7} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = 5\\\frac{{x + 3}}{{\sqrt {x + 4} + 3}} + \frac{{x + 9}}{{\sqrt {x + 11} + 4}} = x + 7\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow x + 7 – \frac{{x + 3}}{{\sqrt {x + 4} + 3}} – \frac{{x + 9}}{{\sqrt {x + 11} + 4}} = 0\\ \Leftrightarrow 2x + 14 – \frac{{2x + 6}}{{\sqrt {x + 4} + 3}} – \frac{{2x + 18}}{{\sqrt {x + 11} + 4}} = 0\\ \Leftrightarrow \left[ {\left( {x + 4} \right) – \frac{{2x + 6}}{{\sqrt {x + 4} + 3}}} \right] + \left[ {\left( {x + 10} \right) – \frac{{2x + 18}}{{\sqrt {x + 11} + 4}}} \right] = 0\\ \Leftrightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + 3x + 12 – 2x – 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 4x + 40 – 2x – 18}}{{\sqrt {x + 11} + 4}} = 0\\ \Leftrightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + x + 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 2x + 22}}{{\sqrt {x + 11} + 4}} = 0\\x \ge – 4 \Rightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + x + 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 2x + 22}}{{\sqrt {x + 11} + 4}} > 0\end{array}\) Suy ra phương trình (1) vô nghiệm Vậy \(x = 5\) Bình luận
Đáp án:
\[x = 5\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge – 4\)
Ta có:
\(\begin{array}{l}
\left( {x + 3} \right)\sqrt {x + 4} + \left( {x + 9} \right)\sqrt {x + 11} = {x^2} + 9x + 10\\
\Leftrightarrow \left( {x + 3} \right).\left( {\sqrt {x + 4} – 3} \right) + \left( {x + 9} \right)\left( {\sqrt {x + 11} – 4} \right) = {x^2} + 9x + 10 – 3\left( {x + 3} \right) – 4\left( {x + 9} \right)\\
\Leftrightarrow \left( {x + 3} \right).\frac{{x + 4 – {3^2}}}{{\sqrt {x + 4} + 3}} + \left( {x + 9} \right).\frac{{x + 11 – {4^2}}}{{\sqrt {x + 11} + 4}} = {x^2} + 2x – 35\\
\Leftrightarrow \frac{{\left( {x + 3} \right)\left( {x – 5} \right)}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 9} \right)\left( {x – 5} \right)}}{{\sqrt {x + 11} + 4}} = \left( {x – 5} \right)\left( {x + 7} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
\frac{{x + 3}}{{\sqrt {x + 4} + 3}} + \frac{{x + 9}}{{\sqrt {x + 11} + 4}} = x + 7\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow x + 7 – \frac{{x + 3}}{{\sqrt {x + 4} + 3}} – \frac{{x + 9}}{{\sqrt {x + 11} + 4}} = 0\\
\Leftrightarrow 2x + 14 – \frac{{2x + 6}}{{\sqrt {x + 4} + 3}} – \frac{{2x + 18}}{{\sqrt {x + 11} + 4}} = 0\\
\Leftrightarrow \left[ {\left( {x + 4} \right) – \frac{{2x + 6}}{{\sqrt {x + 4} + 3}}} \right] + \left[ {\left( {x + 10} \right) – \frac{{2x + 18}}{{\sqrt {x + 11} + 4}}} \right] = 0\\
\Leftrightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + 3x + 12 – 2x – 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 4x + 40 – 2x – 18}}{{\sqrt {x + 11} + 4}} = 0\\
\Leftrightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + x + 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 2x + 22}}{{\sqrt {x + 11} + 4}} = 0\\
x \ge – 4 \Rightarrow \frac{{\left( {x + 4} \right)\sqrt {x + 4} + x + 6}}{{\sqrt {x + 4} + 3}} + \frac{{\left( {x + 10} \right)\sqrt {x + 11} + 2x + 22}}{{\sqrt {x + 11} + 4}} > 0
\end{array}\)
Suy ra phương trình (1) vô nghiệm
Vậy \(x = 5\)
Đáp án:
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Giải thích các bước giải: