Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a+b\sqrt{3}+c\sqrt{5})^2\leq (a^2+b^2+c^2)(1+3+5)\)

\(\Leftrightarrow (a+b\sqrt{3}+c\sqrt{5})^2\leq 9\Rightarrow a+b\sqrt{3}+c\sqrt{5}\leq 3\)

(đpcm)

Dấu "=" xảy ra khi \(\frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{\sqrt{5}}\) hay \(a=\frac{1}{3}; b=\sqrt{\frac{1}{3}}; c=\sqrt{\frac{5}{9}}\)

1) Phương trình đã cho tương đương

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\frac{3}{4}\end{matrix}\right.\)

a) Cả hai phương trình đều có chung \(\sqrt{x+3}\)

pt đầu suy ra  \(\sqrt{x+3}=2\sqrt{y-1}\)

pt sau suy ra \(\sqrt{x+3}=4-\sqrt{y+1}\)

Vậy \(2\sqrt{y-1}=4-\sqrt{y+1}\), đk y > 1

\(4\left(y-1\right)=16-8\sqrt{y+1}+y+1\)

\(8\sqrt{y+1}+3y-21=0\)

Đặt \(\sqrt{y+1}=t\)

=> y = t² - 1

=> 8t + 3(t² -1) -21 =0

3t² + 8t - 24 = 0

=> t = ...

=> y = t² - 1

=> \(\sqrt{x+3}=2\sqrt{y-1}\)

=> x =...

b) Trừ hai pt cho nhau ta có:

x² - y² = 3(y - x)

(x - y) (x + y + 3) = 0

=> x = y hoặc x + y + 3 = 0

Xét hai trường hợp, rút x theo y rồi thay trở lại một trong hai pt ban đầu tìm ra nghiệm

đặt \(\sqrt{x^2+x+1}=t\left(t\ge\sqrt{\dfrac{3}{4}}\right)tacó\)

pt \(\Leftrightarrow\)3t=t\(^2\)+2

\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=2\left(tm\right)\end{matrix}\right.\)

Với t=1 ta có x\(^2\)+x+1=1 \(\Leftrightarrow\)x=0 hoặc x=-1

với t=2 ta có x\(^2\)+x+1 =2 \(\Leftrightarrow\)\(\dfrac{-1\mp\sqrt{5}}{2}\)=x

câu 2 tương tự đặt 2x^2+x-2=t(t\(\ge\dfrac{-17}{8}\))

ta có pt \(\Leftrightarrow\)t^2+5t-6=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=-6\left(loại\right)\end{matrix}\right.\)

với t=1 thì 2x^2+x-2=1 \(\Leftrightarrow\)t=1 hoặc -3/2

\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)

Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)

\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)

4)\(ĐK:x\ge-\dfrac{1}{3}\)

\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)

Vậy pt có 2 nghiệm là x=1 và x=5