\(ĐKXĐ:x\ge-1\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}\) ta được 

\(a^2-a-20=0\Rightarrow\orbr{\begin{cases}a=5\\a=-4\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}-2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{2}{\sqrt{2x+3}+3}+\frac{1}{\sqrt{x+1}+2}\right)=0\)

\(\Rightarrow x=3\)

b ) \(ĐKXĐ:x\ge0\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x}+2x+1+2\sqrt{x^2+x}-2=0\)

Đặt \(\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}\)

\(\Rightarrow a+a^2-2=0\Rightarrow\orbr{\begin{cases}a=1\\a=-2\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{x+1}+\sqrt{x}=1\)

Mà \(x\ge0\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\sqrt{x+1}\ge1\end{cases}\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1}\)

Dấu " = " xảy ra khi và chỉ khi \(x=0\)

áp dụng bất đẳng thức bu nhi a 

ta có \(3\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\)

lại có  a/b+b/c+c/a \(\ge\)3 (bđt cauchy)

nhân từng vế ta có \(3\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{a}+\frac{a}{c}\right)\ge3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\)

suy ra đpcm

Áp dụng định lí Bezout :

\(P\left(-2\right)=-1\Rightarrow4a-2b+3=-1\Rightarrow4a-2b=-4\)

\(P\left(1\right)=8\Rightarrow a+b+3=8\Rightarrow a+b=5\)

\(\Rightarrow\hept{\begin{cases}4a-2b=-4\\a+b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=4\end{cases}}}\)