Question

giải pt :

a)\(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)

b) \(\sqrt{25-x^2}-\sqrt{9-x^2}=2\)

Answer 1

b) ta có pt \(\sqrt{25-x^2}-\sqrt{9-x^2}=2\)

Đặt \(\sqrt{25-x^2}=a;\sqrt{9-x^2}=b\left(a,b\ge0\right)\Rightarrow a-b=2\)

Mà \(a^2-b^2=25-x^2-9+x^2=16\Leftrightarrow\left(a-b\right)\left(a+b\right)=16\Leftrightarrow a+b=8\)

ta có a-b=2;a+b=8=> a=5;b=3

Answer 2

a) ta có pt \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\Leftrightarrow x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)

đặt \(\sqrt{2x-\dfrac{5}{x}}=a;\sqrt{x-\dfrac{1}{x}}=b\Rightarrow a^2-b^2=2x-\dfrac{5}{x}-x+\dfrac{1}{x}=x-\dfrac{4}{x}\)

nên pt \(\Leftrightarrow a^2-b^2+a-b=0\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)