Question

Cho A=\(\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)với \(x\)\(\ge\)\(0\) và \(x\)\(\ne\)\(4\)

a) Rút gọn A

b)Tính giá trị của A trong trường hợp

TH1: \(16x^2-625=0\)

TH2: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

Answer 1

a: \(A=\dfrac{\sqrt{x}-2+7}{x-4}:\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\)

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

b: 

TH1: \(16x^2-625=0\)

=>(4x-25)(4x+25)=0

=>x=25/4

Thay x=25/4 vào A, ta được: 

\(A=\left(\dfrac{5}{2}+5\right):\left(\dfrac{5}{2}+2\right)=\dfrac{15}{2}:\dfrac{9}{2}=\dfrac{5}{3}\)

Th2: \(x=5+\sqrt{2}-4-\sqrt{2}=1\)

Thay x=1 vào A, ta được: \(A=\dfrac{1+5}{1+2}=\dfrac{6}{3}=2\)

Answer 2

a)

A= \(\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)

A= \(\left(\dfrac{\sqrt{x}-2+7}{x-4}\right)\left(\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)

A= \(\dfrac{\sqrt{x}+5}{x-4}\).\(\left(\sqrt{x}-2\right)\)

A = \(\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

b) 

TH1: 16x² - 625 = 0 

<=> (4x - 25).(4x+25) = 0 

<=> \(\left[{}\begin{matrix}x=\dfrac{25}{4}\\x=-\dfrac{25}{4}\end{matrix}\right.\)

do x nằm trong dấu căn nên x = \(-\dfrac{25}{4}\) loại 

=> x = \(\dfrac{25}{4}\) => \(\sqrt{x}\) = \(\dfrac{5}{2}\)

=> A=\(\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\) = \(\dfrac{\dfrac{5}{2}-5}{\dfrac{5}{2}+2}\) = \(\dfrac{15}{2}\).\(\dfrac{2}{9}\) = \(\dfrac{5}{3}\)

TH2: x = \(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

x = \(\sqrt{25+2.5.\sqrt{2}+2}-\sqrt{16+2.4.\sqrt{2}+2}\)

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

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

x = 1 

thay vào A ta có 

A = \(\dfrac{1+5}{1+2}=\dfrac{6}{3}=2\)