Bài 1:Với \(ab=1;a+b\ne0\) ta có: 

\(P=\frac{a^3+b^3}{\left(a+b\right)^3\left(ab\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4\left(ab\right)^2}+\frac{6\left(a+b\right)}{\left(a+b\right)^5\left(ab\right)}\)

\(=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)

\(=\frac{a^2+b^2-1}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)

\(=\frac{\left(a^2+b^2-1\right)\left(a+b\right)^2+3\left(a^2+b^2\right)+6}{\left(a+b\right)^4}\)

\(=\frac{\left(a^2+b^2-1\right)\left(a^2+b^2+2\right)+3\left(a^2+b^2\right)+6}{\left(a+b\right)^4}\)

\(=\frac{\left(a^2+b^2\right)^2+4\left(a^2+b^2\right)+4}{\left(a+b\right)^4}=\frac{\left(a^2+b^2+2\right)^2}{\left(a+b\right)^4}\)

\(=\frac{\left(a^2+b^2+2ab\right)^2}{\left(a+b\right)^4}=\frac{\left[\left(a+b\right)^2\right]^2}{\left(a+b\right)^4}=1\)

Bài 2: \(2x^2+x+3=3x\sqrt{x+3}\)

Đk:\(x\ge-3\)

\(pt\Leftrightarrow2x^2-3x\sqrt{x+3}+\sqrt{\left(x+3\right)^2}=0\)

\(\Leftrightarrow2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\sqrt{\left(x+3\right)^2}=0\)

\(\Leftrightarrow2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x+3}\right)\left(2x-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=x\\\sqrt{x+3}=2x\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\ge0\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-x-3=0\left(x\ge0\right)\\4x^2-x-3=0\left(x\ge0\right)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{13}}{2}\\x=1\end{cases}\left(x\ge0\right)}\)

Bài 4:

Áp dụng BĐT AM-GM ta có: 

\(2\sqrt{ab}\le a+b\le1\Rightarrow b\le\frac{1}{4a}\)

Ta có: \(a^2-\frac{3}{4a}-\frac{a}{b}\le a^2-\frac{3}{4a}-4a^2=-\left(3a^2+\frac{3}{4a}\right)\)

\(=-\left(3a^2+\frac{3}{8a}+\frac{3}{8a}\right)\le-3\sqrt[3]{3a^2\cdot\frac{3}{8a}\cdot\frac{3}{8a}}=-\frac{9}{4}\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{2}\)

\(gt\Rightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(\Rightarrow\frac{1}{a^2}+1=\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\)

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

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

\(=\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{bc}+\frac{1}{ca}\)

Tương tự với các cụm còn lại, ta được

\(A=2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)

bài này khó thật, nhưng bạn đừng buồn, sẽ có nhiều bạn khác giúp bạn

nha Nguyễn Quang Linh à

chắc =1 đó

\(A=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{\left(ab\right)^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{\left(ab\right)^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{ab}\)

\(=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{1^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{1^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{1}\)

\(=\frac{a^2-ab+b^2}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)\(=\frac{\left(a^3+b^3\right)\left(a+b\right)+3a^2+3b^2+6}{\left(a+b\right)^4}\)

\(=\frac{a^4+a^3b+ab^3+b^4+3a^2+3b^2+6}{a^4+4a^3b+6a^2b^2+4ab^3+b^4}\)\(=\frac{a^4+a^2.1+1.b^2+b^4+3a^2+3b^2+6}{a^4+4a^2.1+6.1^2+4b^2.1+b^4}\)

\(=\frac{a^4+4a^2+4b^2+b^4+6}{a^4+4a^2+6+4b^2+b^4}=1\)