Question

giải pt  \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{2x-3}}=\sqrt{3}\left(\frac{1}{\sqrt{4x-3}}+\frac{1}{\sqrt{5x-6}}\right)\)

Answer 1

\(\frac{1}{pt}\)=\(\sqrt{x}+\sqrt{2x+3}=\frac{1}{\sqrt{3}}\left(\sqrt{4x-3}+\sqrt{5x-6}\right)\)   

=>\(\frac{x-2x-3}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{4x-3-5x-6}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)

=>\(\frac{3-x}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{3-x}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)

=>\(\sqrt{x}-\sqrt{2x-3}=\sqrt{3}\left(\sqrt{4x-3}-\sqrt{5x+6}\right)\)

=>\(\frac{3-x}{\sqrt{x}+\sqrt{2x-3}}=\sqrt{3}\left(\frac{3-x}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)

=>\(\left(3-x\right)\left(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\right)\)=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

Answer 2

Em mới học lớp 7 

Answer 3

x=3 nha

Answer 4

uk nghiệm = 3 nhưng cách giải ntn 

Answer 5

e moi hoc lop 7

Answer 6

\(\sqrt{3}\left(\sqrt{x}+\sqrt{2x-3}\right)=\sqrt{4x-3}+\sqrt{5x-6}\)

Giải cái này cũng ra x=3

Answer 7

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