\(2\sqrt{3}-\sqrt{27}+\sqrt{4-2\sqrt{3}}\)

Ta có : \(4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow2\sqrt{3}-3\sqrt{3}+\sqrt{3}-1\)vì \(\sqrt{3}-1>0\)

\(=-1\)Vậy biểu thức trên nhận giá trị là -1 

ĐK x > \(\frac{2}{3}\)

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

\(\Leftrightarrow3x-2=x^4-4x^3+4x^2+4x^2-8x+4\)\(\Leftrightarrow x^4-4x^3+8x^2-11x+6=0\)

\(\Leftrightarrow x^4-x^3-3x^3+3x^2+5x^2-5x-6x+6=0\)

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)+5x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2+5x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x^3-3x^2+5x-6=0\end{cases}}\)\(\hept{\left(1;2\right)}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\\left(x-2\right)\left(x^2-x+3\right)=0\end{cases}}\)

Vậy x=1,x=2

\(\orbr{\begin{cases}x=1\\\orbr{\begin{cases}x=2\\x^2-x+3=0\end{cases}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\\orbr{\begin{cases}x=2\left(tm\right)\\x^2-x+3=0\left(loai\right)\end{cases}}\end{cases}}\)

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

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

\(\Leftrightarrow\frac{3x-6}{\sqrt{3x-2}+2}-x\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\frac{3}{\sqrt{3x-2}+2}=x\left(1\right)\end{cases}}\)

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

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

\(\Leftrightarrow x^2\left(3x-2\right)=9+4x^2-12x\)

\(\Leftrightarrow3x^3-2x^2=9+4x^2-12x\)

\(\Leftrightarrow3x^3-6x^2+12x-9=0\)

\(\Leftrightarrow x^3-2x^2+4x-3=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x^2-x+3=0\left(2\right)\end{cases}}\)

\(\Delta_{\left(2\right)=1^2-3.4=-11< 0}\)( vô nghiệm )

Vậy pt có tập nghiệm \(S=\left\{1;2\right\}\)

Để pt có 2 nghiệm phân biệt thì \(\Delta>0\)

hay \(\left[2\left(m-1\right)\right]^2-4\left(m+1\right)=\left(2m-2\right)^2-4m-4\)

\(=4m^2-8m+4-4m-4=4m^2-12m>0\)

\(\Leftrightarrow4m\left(m-3\right)>0\)

TH1 : \(\hept{\begin{cases}4m>0\\m-3>0\end{cases}\Leftrightarrow\hept{\begin{cases}m>0\\m>3\end{cases}\Leftrightarrow m>3}}\)

TH2 : \(\hept{\begin{cases}4m< 0\\m-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}m< 0\\m< 3\end{cases}\Leftrightarrow m< 0}}\)

Vậy với m > 3 ; m < 0 thì pt có 2 nghiệm pb 

ta có:

denta= b² - 4ac =(m-1)² - 4(m+1).1= m² - 6m - 3

để phương trình có 2 no pb thì denta > 0

=> m² - 6m - 3 > 0  \(\Leftrightarrow\orbr{\begin{cases}x< 3-2\sqrt{3}\\3+2\sqrt{3}< x\end{cases}}\)

câu a là cmr tứ giác PHIB nội tếp

giải pt (1) ta có:

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

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

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

=> x=3y+1 thay vào (2) => x=1; y=0

trường hợp 2:

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

=> \(\sqrt{3y+1}+\sqrt{x}\)=\(\sqrt{x+2y}+\sqrt{2x-y-1}\)

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

=> \(\frac{x-2x+y+1}{\sqrt{x}+\sqrt{2x-y-1}}\)+\(\frac{3y+1-x-2y}{\sqrt{3y+1}+\sqrt{x+2y}}\)=0

=>(-x + y + 1)(\(\frac{1}{\sqrt{x}+\sqrt{2x-y-1}}\)+ \(\frac{1}{\sqrt{3y+1}+\sqrt{x+2y}}\))=0

mà \(\frac{1}{\sqrt{x}+\sqrt{2x-y-1}}\)+\(\frac{1}{\sqrt{3y+1}+\sqrt{x+2y}}\)>0

=> x=y+1 thay vào 2 => \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)

để đấy ku

\(\frac{a^2+2}{\sqrt{a^2+1}}=\frac{a^2+1+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\frac{1}{\sqrt{a^2+1}}\ge2\)

\(\forall a\inℝ\)

ta có: a² + 2 \(\ge\)\(2\sqrt{a^2+1}\)

\(\Rightarrow\)a² + 1 -\(2\sqrt{a^2+1}\)+ 1 \(\ge\)0

\(\Rightarrow\)(\(\sqrt{a^2+1}\)-  1)² \(\ge\)0 (luôn đúng)