\(M=\left(\frac{3}{\sqrt{a+1}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}+1\right)\)

\(=\left(\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\right):\left(\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1+a\right)\left(1-a\right)}}\right)\)

\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}.\frac{\sqrt{\left(1+a\right)\left(1-a\right)}}{3+\sqrt{\left(1-a\right)\left(1+a\right)}}\)

\(=\sqrt{1-a}\left(đpcm\right)\)

bn viết lại pt đi

đây nè bạn \(A=\left(\sqrt{a}+\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\\a\ne4\end{matrix}\right.\)

\(Q=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\frac{\sqrt{a}-2}{3\sqrt{a}}\)

b) Vì \(3\sqrt{a}\ge0\forall a\ge0\)nên để Q dương thì

\(\left\{{}\begin{matrix}3\sqrt{a}>0\\\sqrt{a}-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\a>4\end{matrix}\right.\Leftrightarrow a>4\)

c) Ta có: \(\sqrt{a}=\sqrt{9-4\sqrt{5}}=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}-3}=\frac{\sqrt{5}-2-2}{3\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-4}{3\left(\sqrt{5}-2\right)}=\frac{-3-2\sqrt{5}}{3}\)

(cái cuối nhân lượng liên hợp để ra kết quả cuối cùng)

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:

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