Question

[Ariels spring fashion]

BT1: Rút gọn

A = \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\).\(\frac{\sqrt{a}+1}{\sqrt{a}}\)

B = \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\)\(\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

BT2: Cho A = \(\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right)\)\(\frac{\left(1-a\right)^2}{2}\)

a) Rút gọn A

b) Chứng minh rằng: Nếu 0<a<1 thì A >0

c) Tính giá trị lớn nhất của a

BT3: Rút gọn

C = \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)- \(\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

BT4: Tìm giá trị nhỏ nhất

      H = \(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\)

          Tìm giá trị lớn nhất

     G = \(\frac{x-5}{\sqrt{x-1}-\sqrt{2}}-x+2\)

Answer 1

[Đoàn Đức Hà]

BT1: 

ĐK: \(a>0,a\ne1\).

\(A=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{a+\sqrt{a}-2-\left(a-\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{2}{a-1}\)

ĐK: \(a\ge0,a\ne1\).

\(B=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(B=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(B=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

Answer 2

[Đoàn Đức Hà]

BT2: 

a) ĐK: \(a\ge0,a\ne1\).

\(A=\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\left(\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{a-\sqrt{a}-2-\left(a+\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{-2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)^2}{2}\)

\(A=-\sqrt{a}\left(\sqrt{a}-1\right)=\sqrt{a}-a\)

b) \(A=\sqrt{a}-a=\sqrt{a}\left(1-\sqrt{a}\right)>0\)

(Vì \(0< a< 1\Rightarrow1-\sqrt{a}>0\))

c) \(A=\sqrt{a}-a=\frac{1}{4}-\left(\frac{1}{4}-\sqrt{a}+a\right)=\frac{1}{4}-\left(\sqrt{a}-\frac{1}{2}\right)^2\le\frac{1}{4}\)

Dấu \(=\)xảy ra khi \(\sqrt{a}-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{4}\). Vậy GTLN của \(A\) là \(\frac{1}{4}\).