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

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

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

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

=1

#mã mã#

B ơi b lấy đề này ở đâu v ạ

Áp dụng BĐT bunniacoxki ta có:

\(\left(b^2+\left(c+a\right)^2\right)\left(1+4\right)\ge\left(b+2\left(a+c\right)\right)^2\)

=> \(\sqrt{\frac{a^2}{b^2+\left(c+a\right)^2}}\le\sqrt{5}.\frac{a}{b+2c+2a}\)

=> \(VT\le\sqrt{5}.\left(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\right)\)

Cần CM \(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\le\frac{3}{5}\)

<=>\(\left(\frac{1}{2}-\frac{a}{b+2c+2a}\right)+\left(\frac{1}{2}-\frac{b}{c+2a+2b}\right)+\left(\frac{1}{2}-\frac{c}{a+2b+2c}\right)\ge\frac{9}{10}\)

<=>\(\frac{b+2c}{b+2c+2a}+\frac{c+2a}{c+2a+2b}+\frac{a+2b}{a+2b+2c}\ge\frac{9}{5}\)

Áp dụng bđt buniacoxki dạng phân thức ở vế trái:

=> \(VT\ge\frac{\left(b+2c+c+2a+a+2b\right)^2}{\left(b+2c\right)^2+2a\left(b+2c\right)+\left(c+2a\right)^2+2b\left(c+2a\right)+\left(a+2b\right)^2+2c\left(a+2b\right)}\)

         \(=\frac{9\left(a+b+c\right)^2}{5\left(a+b+c\right)^2}=\frac{9}{5}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

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

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

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

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

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

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

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

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

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

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

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

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

a) ĐKXĐ: x\(\ne\) 0;4

Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

                   = \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

                   =\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)

b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)

<=> \(4=3-\sqrt{x}\)

<=> \(\sqrt{x}=-1\) (vô lí)

Vậy ko tìm được x.

Kết quả câu a như nào nhỉ ?

a) đk: \(\hept{\begin{cases}a>b\\a< -b\end{cases}}\left(b>0\right)\) hoặc \(\hept{\begin{cases}a>-b\\a< b\end{cases}\left(b< 0\right)}\)

Ta có:
\(B=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right)\div\frac{b}{a-\sqrt{a^2-b^2}}\)

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

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

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

b) \(B< 1\Leftrightarrow\sqrt{\frac{a-b}{a+b}}< 1\Leftrightarrow\frac{a-b}{a+b}< 1\)

\(\Leftrightarrow\frac{-2b}{a+b}< 0\) ta xét 2TH:

Nếu \(b>0\Rightarrow a>-b\)

Nếu \(b< 0\Rightarrow a< -b\)

Vậy ...