Question

[Thijnh]

bài 5.6.8

làm hộ mình nhé

Answer 1

[Đoàn Đức Hà]

ĐK: \(x\ge0,x\ne4\).

Với \(x=7-4\sqrt{3}\):

\(\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

\(B=\frac{2}{2-\sqrt{3}-2}=\frac{-2}{\sqrt{3}}=\frac{-2\sqrt{3}}{3}\)

\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

\(P=\frac{4}{3}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{4}{3}\Leftrightarrow3\left(\sqrt{x}+2\right)=4\left(\sqrt{x}+1\right)\Leftrightarrow x=4\)(loại) \(\left(\sqrt{x}+1\right)P-\sqrt{x}-4\sqrt{x-1}+26=-6+10\sqrt{5x}\)

\(\Rightarrow\sqrt{x}+2-\sqrt{x}-4\sqrt{x-1}+26+6x-10\sqrt{5x}=0\)

\(\Leftrightarrow x+3-4\sqrt{x-1}+5x+25-10\sqrt{5x}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)^2-4\left(x-1\right)}{x+3+4\sqrt{x-1}}+\frac{5\left[\left(x+5\right)^2-\left(2\sqrt{5x}\right)^2\right]}{x+5+2\sqrt{5x}}=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)\left(\frac{1}{x+3+4\sqrt{x-1}}+\frac{5}{x+5+2\sqrt{5x}}\right)=0\)

\(\Leftrightarrow x^2-10x+25=0\)(vì \(x\ge0\)) 

\(\Leftrightarrow x=5\)(thỏa mãn)