rut gon bieu thuc x+2tren xcanx-1 + can x =1 tren x +can x +1 20/11/2021 Bởi Faith rut gon bieu thuc x+2tren xcanx-1 + can x =1 tren x +can x +1
Đáp án: $\begin{array}{l}Dkxd:x \ge 0;x \ne 1\\\frac{{x + 2}}{{x\sqrt x – 1}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\ \Rightarrow \frac{{x + 2}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\ \Rightarrow \frac{{x + 2 + \sqrt x \left( {x\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{\sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}\\ \Rightarrow x + 2 + {x^2} – \sqrt x = \sqrt x – 1\\ \Rightarrow {x^2} + x – 2\sqrt x + 3 = 0\\ \Rightarrow {x^2} + \left( {x – 2\sqrt x + 1} \right) + 2 = 0\\ \Rightarrow {x^2} + {\left( {\sqrt x – 1} \right)^2} + 2 = 0\left( {vô\,nghiệm} \right)\end{array}$ Vậy phương trình vô nghiệm. Bình luận
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
\frac{{x + 2}}{{x\sqrt x – 1}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\
\Rightarrow \frac{{x + 2}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\
\Rightarrow \frac{{x + 2 + \sqrt x \left( {x\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{\sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
\Rightarrow x + 2 + {x^2} – \sqrt x = \sqrt x – 1\\
\Rightarrow {x^2} + x – 2\sqrt x + 3 = 0\\
\Rightarrow {x^2} + \left( {x – 2\sqrt x + 1} \right) + 2 = 0\\
\Rightarrow {x^2} + {\left( {\sqrt x – 1} \right)^2} + 2 = 0\left( {vô\,nghiệm} \right)
\end{array}$
Vậy phương trình vô nghiệm.