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

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

\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)

\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)

\(A=\frac{a-b}{\sqrt{a-b}.\sqrt{a+b}}\)

\(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)

Với \(a=3b\) ta có : \(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}}{2}\)

Chúc bạn học tốt ~ 

mn làm giúp mk vs

\(\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{\sqrt{a^2-b^2}b}\)

\(=\frac{ab-a^2+a^2-b^2}{\sqrt{a^2-b^2}b}\)

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

\(=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)

b, Thay a = 3b

\(=\sqrt{\frac{3b-b}{3b+b}}=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}\)

bài này cũng tương tự câu trên vậy tách màu ra là tính được mà . đâu có khó gì đâu bạn . 

Biến đổi vế trái :vvv

\(VT=\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)

\(=\frac{a+b}{b^2}.\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)

\(=\frac{a+b}{b^2}.\frac{\left|ab^2\right|}{\left|a+b\right|}\)

\(=\frac{a+b}{b^2}.\frac{b^2.\left|a\right|}{a+b}=\left|a\right|=VP\left(đpcm\right)\)

( Vì a + b > 0 nên | a + b | = a + b ; b² > 0 )

Rút gọn bt:

Câu 1: a, \(\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)

b, \(\sqrt{25a}+2\sqrt{45a}-3\sqrt{80a}+2\sqrt{16a}\left(a\ge0\right)\)ư

Câu 2: Cho bt: P =\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

a, Tìm ĐKXĐ . Rút gọn P 

B, Tìm x nguyên để P có gt nguyên

c, Tìm GTNN của P với a >1

Câu 3: Giair các pt 

a, \(\sqrt{\left(2x-1\right)^2}=4\)

b, \(\sqrt{4x+4}+\sqrt{9x+9}-8\sqrt{\frac{x+1}{16}}=5\)

Bài 1:

\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

Bài 2:

\(a,P=\left(\frac{1}{x+\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+2\sqrt{x}+1}\left(x>0\right)\)

\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\times\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

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

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

\(=\frac{1-x}{x}\)

\(b,\forall x>0\Leftrightarrow\frac{1-x}{x}>\frac{1}{2}\)

\(\Leftrightarrow2\left(1-x\right)>x\)

\(\Leftrightarrow2-2x>x\)

\(\Leftrightarrow-3x>-2\)

\(\Leftrightarrow x< \frac{2}{3}\)

\(\Rightarrow P>\frac{1}{2}\Leftrightarrow\forall0< x< \frac{2}{3}\)