Giải pt:
a) x=\(\sqrt{1-\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}\)
b) \(\sqrt{x^2+x}+\sqrt{x-x^2}=x+1\)
c) \(\sqrt{x^2-x}+\sqrt{x^2+2x}=2\sqrt{x^2}\)
d)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
e) \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)
f)...
Đọc tiếp
Giải pt:
a) x=\(\sqrt{1-\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}\)
b) \(\sqrt{x^2+x}+\sqrt{x-x^2}=x+1\)
c) \(\sqrt{x^2-x}+\sqrt{x^2+2x}=2\sqrt{x^2}\)
d)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
e) \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)
f) \(4x\sqrt{x+7}+3x\sqrt{7x-3}=6x^2+2\sqrt{7x^2+46x-21}\)
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)