\(a-b=\sqrt{1-b^2}-\sqrt{1-a^2}\Leftrightarrow a+\sqrt{1+a^2}=b+\sqrt{1+b^2}\)

Bình phương cả 2 vế: \(2a\sqrt{1+a^2}=2b\sqrt{1+b^2}\)

Tiếp tục bình phương: \(a^2+a^4=b^2+b^4\)

\(\Leftrightarrow a^2-b^2+\left(b^2-a^2\right)\left(a^2+b^2\right)=0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(1-a^2-b^2\right)=0\)

Đến đây ta có: \(\orbr{\begin{cases}a=b\\a^2+b^2=1\end{cases}}\)

Nếu a=b sẽ có vô số a,b TMDK nên đề bài nên có thêm điều kiện a,b phân biệt

\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)

\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)

\(=\dfrac{11}{a-9}\)

\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)

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

\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(\text{đ}k\text{x}\text{đ}:a\ge0;a\ne9\right)\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a-3}\right)\left(\sqrt{a+3}\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\dfrac{a-2}{\left(\sqrt{a}+3\right)\left(\sqrt{a-3}\right)}\\ =\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{11}{\left(\sqrt{a}-3\right)\left(\sqrt{a+3}\right)}\)

\(b,\dfrac{a+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}\\ =\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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

Bài 1:

Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)

xong bn áp dụng lên trên lm tiếp

Bài 3:

theo bđt cô si ta có:

\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)

=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)                         (1)

Tương tự ta có :

\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\)                            (2)

\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)                               (3)

Cộng vế vs vế (1)(2)(3) ta có:

\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)

Bài 2:

Ta có: 

\(\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Nên:

\(A< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\right)=\frac{1}{2}\left(1-\frac{1}{5}\right)=\frac{2}{5}\)

Lời giải:

a) ĐK: $a>0; a\neq 1$

b)

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

\(=\frac{\sqrt{a}+2}{\sqrt{a}(\sqrt{a}+1)}-\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{\sqrt{a}(a-1)}=\frac{(\sqrt{a}+2)(\sqrt{a}-1)}{(a-1)\sqrt{a}}-\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{\sqrt{a}(a-1)}\)

\(=\frac{(a+\sqrt{a}-2)-(a-\sqrt{a}-2)}{(a-1)\sqrt{a}}=\frac{2\sqrt{a}}{\sqrt{a}(a-1)}=\frac{2}{a-1}\) (đpcm)

a) Trục căn thức ở mỗi số hạng của biểu thức A,ta có:

 \(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)=\(\frac{\sqrt{2}+\sqrt{1}}{1-2}-\frac{\sqrt{3}+\sqrt{2}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...+\frac{\sqrt{2007}+\sqrt{2008}}{2007-2008}\)

= \(-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...-\left(\sqrt{2007}+\sqrt{2008}\right)\)

=\(-1-\sqrt{2008}\)

b)Ta xét số hạng tổng quát: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)=\(\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)=\(\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào biểu thức B ta được: 

B= \(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-...+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}\)= \(\frac{10}{11}\)

\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)

\(=\frac{-1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{1}{\sqrt{4}-\sqrt{3}}+\frac{1}{\sqrt{5}-\sqrt{4}}-....+\frac{1}{\sqrt{2007}-\sqrt{2006}}-\frac{1}{\sqrt{2008}-\sqrt{2007}}\)

\(=\frac{-1\cdot\left(\sqrt{2}+\sqrt{1}\right)}{2-1}+\frac{1\cdot\left(\sqrt{3}+\sqrt{2}\right)}{3-2}-\frac{1\cdot\left(\sqrt{4}+\sqrt{3}\right)}{4-3}+\frac{1\cdot\left(\sqrt{5}+\sqrt{4}\right)}{5-4}-...+\frac{1\cdot\left(\sqrt{2007}+\sqrt{2006}\right)}{2007-2006}-\frac{1 \left(\sqrt{2008}+\sqrt{2007}\right)}{2008-2007}\)

\(=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-...+\sqrt{2006}+\sqrt{2007}-\sqrt{2007}-\sqrt{2008}\) 

\(=-1-\sqrt{2008}\)

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