\(\frac{1}{pt}\)=\(\sqrt{x}+\sqrt{2x+3}=\frac{1}{\sqrt{3}}\left(\sqrt{4x-3}+\sqrt{5x-6}\right)\)
=>\(\frac{x-2x-3}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{4x-3-5x-6}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)
=>\(\frac{3-x}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{3-x}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)
=>\(\sqrt{x}-\sqrt{2x-3}=\sqrt{3}\left(\sqrt{4x-3}-\sqrt{5x+6}\right)\)
=>\(\frac{3-x}{\sqrt{x}+\sqrt{2x-3}}=\sqrt{3}\left(\frac{3-x}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)
=>\(\left(3-x\right)\left(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\right)\)=0
=>3-x=0=>x=3
hoặc\(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)=0
\(\sqrt{3}\left(\sqrt{x}+\sqrt{2x-3}\right)=\sqrt{4x-3}+\sqrt{5x-6}\)
Giải cái này cũng ra x=3
Ta có: \(\frac{1}{\sqrt{x}}-\frac{\sqrt{3}}{\sqrt{4x-3}}=\frac{\sqrt{3}}{\sqrt{5x-6}}-\frac{1}{\sqrt{2x-3}}\)
\(\Rightarrow\frac{\sqrt{4x-3}-\sqrt{3x}}{\sqrt{x\left(4x-3\right)}}=\frac{\sqrt{3\left(2x-3\right)}-\sqrt{5x-6}}{\sqrt{\left(5x-6\right)\left(2x-3\right)}}\)
\(\Rightarrow\frac{x-3}{\sqrt{x\left(4x-3\right)}.\left(\sqrt{4x-3}+\sqrt{3x}\right)}=\frac{x-3}{\sqrt{\left(5x-6\right)\left(2x-3\right)}.\left(\sqrt{3\left(2x-3\right)}+\sqrt{5x-6}\right)}\)
Tự làm tiếp