Áp dụng BĐT AM-GM ta có: 

\(\frac{a^2}{b+c}+\frac{b+c}{\frac{9}{4}}+\frac{b^2}{c+a}+\frac{a+c}{\frac{9}{4}}+\frac{16c^2}{a+b}+a+b\)

\(\ge2\sqrt{\frac{a^2}{b+c}\cdot\frac{b+c}{\frac{9}{4}}}+2\sqrt{\frac{b^2}{c+a}\cdot\frac{a+c}{\frac{9}{4}}}+2\sqrt{\frac{16c^2}{a+b}\cdot\left(a+b\right)}=\frac{4a+4b}{3}+8c\)

Suy ra 

\(VT\ge\frac{4a+4b}{3}+8c-\frac{b+c}{\frac{9}{4}}-\frac{c+a}{\frac{9}{4}}-\left(a+b\right)=\frac{64c-a-b}{9}=VP\)

Dấu "=" khi \(a=b=2c\) 

Bài này bạn cũng chú ý tới dấu "=" là xong nhé.

Xét hiệu \(VP-VT=\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)-\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\right)\)

\(=\frac{3a^3b^2+5a^3c^2+3a^2b^3-9a^2b^2c-7a^2bc^2+5a^2c^3+3ab^3c-8ab^2c^2-3abc^3+4b^3c^2+4b^2c^3}{4abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Dễ thấy: \(a;b;c>0\) nên cần chứng minh 

\(3a^3b^2+5a^3c^2+3a^2b^3-9a^2b^2c-7a^2bc^2+5a^2c^3+3ab^3c-8ab^2c^2-3abc^3+4b^3c^2+4b^2c^3\ge0\)

\(\Leftrightarrow\frac{1}{2}\left(8a^3+5a^2b+3a^2c-4ab^2-4ac^2-b^3+3b^2c+5bc^2+c^3\right)\left(b-c\right)^2+\frac{1}{2}\left(3a^2c-2a^3-5a^2b+4ab^2+4ac^2+7b^3+3b^2c-5bc^2-c^3\right)\left(c-a\right)^2+\frac{1}{2}\left(2a^3+5a^2b-3a^2c+4ab^2+4ac^2+b^3-3b^2c+5bc^2+9c^3\right)\left(a-b\right)^2\ge0\)

Tớ ko hiểu lắm

Bạn chú ý tới dấu "=" của BĐT để tìm cách tách hợp lí nhé.

a)Áp dụng BĐT AM-GM ta có:

\(VT=\frac{a^2\cdot2b}{2}\le\frac{\frac{\left(a+a+2b\right)^3}{27}}{2}=\frac{\frac{\left(2\left(a+b\right)\right)^3}{27}}{2}=\frac{4}{27}=VP\)

Dấu "=" khi \(\left(a;b\right)=\left(\frac{2}{3};\frac{1}{3}\right)\)

b)Áp dụng BĐT AM-GM ta có:

\(VT=ab+\frac{1}{ab}=ab+\frac{1}{16ab}+\frac{15}{16ab}\)

\(\ge2\sqrt{ab\cdot\frac{1}{16ab}}+\frac{15}{16\cdot\frac{\left(a+b\right)^2}{4}}\)

\(\ge2\sqrt{ab\cdot\frac{1}{16ab}}+\frac{15}{16\cdot\frac{\left(a+b\right)^2}{4}}\)

Dấu "=" khi \(a=b=\frac{1}{2}\)

\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(\Rightarrow\)\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

                       \(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

                       \(=\sqrt{3}+1+\sqrt{3}-1\)

                       \(=2\sqrt{3}\)

\(\Rightarrow\)\(A=\sqrt{6}\)   (đpcm)

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

\(VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

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

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

\(=\frac{\left|\sqrt{3}+\sqrt{1}\right|+|\sqrt{3}-\sqrt{1}|}{\sqrt{2}}\)

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

\(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\)

\(=VP\)

Vậy đẳng thức được chứng minh .

a) Ta có:  \(AC^2+AB^2=4,5^2+6^2=56,25\)

                  \(BC^2=7,5^2=56,25\)

suy ra:  \(AC^2+AB^2=BC^2\)

hay  tam giác ABC vuông tại A

Áp dụng hệ thức lượng ta có:

\(AH.BC=AB.AC\)

\(\Rightarrow\)\(AH=\frac{AB.AC}{BC}=3,6\)

b)  Áp dụng hệ thức lượng ta có:

   \(AB^2=BH.BC\)

\(\Rightarrow\)\(BH=\frac{AB^2}{BC}=4,8\)

\(\Rightarrow\)\(HC=BC-BH=7,5-4,8=2,7\)

ĐK:  \(x\ge0\);   \(x\ne4\)

\(\frac{-3\sqrt{x}+6}{-x+4\sqrt{x}-4}\)

\(=-\frac{3\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\)

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

\(\frac{-3\sqrt{x}+6}{-x+4\sqrt{x}-4}\)

\(=\frac{3\sqrt{x}-6}{x-4\sqrt{x}+4}\)

\(=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}\)

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