Question

[Đừng gọi tôi là Jung Hae Ri]

Cho biểu thức: \(P=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right):\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)

a, Rút gọn P

b, Tính giá trị của P biết \(x=7-4\sqrt{3}\)

c, Tìm x để P > x.

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: ...

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}-\frac{1-\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\right):\left(\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{1-\sqrt{x}}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x-\sqrt{x}+1}\right)\)

\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1+\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}.\frac{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)

\(\Rightarrow P=\frac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}=\frac{6-3\sqrt{3}}{2-\sqrt{3}}=3\)

Câu c hơi nghi ngờ cái đề, cấp 2 làm sao giải được BPT bậc 3 kiểu này?