2 và 1 + √2
ta có :
1 + √2
= 1,5 + 1 
= 2,5
<=> 2 và 2,5
<=> 2 < 2,5
<=> 2 < 1 + √2

\(\sqrt{3}< \sqrt{4}\)

\(\Rightarrow\sqrt{3}-1< \sqrt{4}-1\)

\(\Rightarrow\sqrt{3}-1< 2-1=1\)

Pt a: Đk \(1< x\le6\)
\(\frac{\sqrt{6-x}-2x+3}{\sqrt{x-1}}=\sqrt{x-1}\Rightarrow\sqrt{6-x}-2x+3=x-1\)
\(\Leftrightarrow\sqrt{6-x}=3x-4\Rightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow6-x=9x^2-24x+16\Leftrightarrow9x^2-23x+10=0\)
\(\Leftrightarrow9x^2-18x-5x+10=0\Leftrightarrow9x\left(x-2\right)-5\left(x-2\right)=0\Leftrightarrow\left(9x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x-5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{9}\left(Lọai\right)\\x=2\left(Thoả\right)\end{cases}}\)
Vậy \(S=\left\{2\right\}\)
Pt b :
Đk: \(x^2-4\ge0\Leftrightarrow x^2\ge4\Leftrightarrow\left|x\right|\ge2\Leftrightarrow\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)
\(\left(x+1\right)\sqrt{x^2-4}=2x+2\Leftrightarrow\left(x+1\right)\left(\sqrt{x^2-4}-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\sqrt{x^2-4}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(Lọai\right)\\\sqrt{x^2-4}=2\end{cases}}\)
\(\Leftrightarrow\sqrt{x^2-4}=2\Rightarrow x^2-4=4\Leftrightarrow x^2=8\Leftrightarrow x=2\sqrt{2}\left(Thoả\right)\)
Vậy \(S=\left\{2\sqrt{2}\right\}\)

\(a,\sqrt{4-4x+x^2}+\sqrt{\frac{2}{x^2+6x+9}}=\sqrt{\left(x-2\right)^2}+\sqrt{\frac{2}{\left(x+3\right)^2}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x+3>0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x>-3\end{cases}\Rightarrow}x\ge-2}\)

\(b,\frac{5\sqrt{x}}{\sqrt{x}-3}+\frac{2}{\sqrt{x}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\sqrt{x}-3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\ne\sqrt{9}\end{cases}\Rightarrow}\hept{\begin{cases}x>0\\x\ne9\end{cases}}}\)

\(c,\sqrt{3-\sqrt{x}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\3-\sqrt{x}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\le3\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\le9\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x\le3\end{cases}}}\)

\(\Rightarrow0< x\le3\)

c,  Để BT có nghĩa thì  \(x^2-4x+3\ge0\)

                                    \(\Leftrightarrow x^2-4x+4\ge1\)

                                    \(\Leftrightarrow\left(x-2\right)^2\ge1\)

                                    \(\Leftrightarrow\sqrt{\left(x-2\right)^2}\ge1\)

                                      \(\Leftrightarrow|x-2|\ge1\)

\(\Leftrightarrow x-2\ge1\) và   \(x-2\le-1\)

\(\Leftrightarrow x\ge3;x\le1\)

Bằng không!

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

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

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

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

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

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

\(2M=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)                  

 để 2M có giá trị nguyên thì \(2\sqrt{x}+2⋮\sqrt{x}+2\)(1)

Lại có \(2\sqrt{x}+4⋮\sqrt{x}+2\)(2)

\(\Rightarrow2⋮\sqrt{x}+2\)(lấy (2) trừ (1))

mà \(\sqrt{x}+2\ge2\)

\(\Rightarrow\sqrt{x}+2=2\)   ( vì x thuộc Z)

=> x=0

Ta có: \(M=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)  ( ĐK: \(x\ge0\) )

\(\Leftrightarrow2M=\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)

\(\Leftrightarrow2M=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)

\(\Leftrightarrow2M=\frac{2\sqrt{x}+4-2}{\sqrt{x}+2}\)

\(\Leftrightarrow2M=\frac{2\sqrt{x}+4}{\sqrt{x}+2}-\frac{2}{\sqrt{x}+2}\)

\(\Leftrightarrow2M=2-\frac{2}{\sqrt{x}+2}\)

Để 2M có giá trị nguyên <=> \(2⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(2\right)\)

\(\Leftrightarrow\sqrt{x}+2\in\left\{-1;-2;1;2\right\}\)

Vì \(x\ge0\Leftrightarrow\sqrt{x}+2\ge2\)

\(\Rightarrow\sqrt{x}+2=2\)

\(\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)

Vậy khi x = 0 thì 2M có giá trị nguyên! 

Chúc bạn học tốt! :))

\(\hept{\begin{cases}x^3-3x-2=2-y\\y^3-3y-2=4-2z\\z^3-3z-2=6-3x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^3-x-2x-2=2-y\\y^3-y-2y-2=2\left(2-z\right)\\z^3-z-2z-2=3\left(2-x\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\left(x^2-1\right)-2\left(x+1\right)=2-y\\y\left(y^2-1\right)-2\left(y+1\right)=2\left(2-z\right)\\z\left(z^2-1\right)-2\left(z+1\right)=3\left(2-x\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left[x\left(x-1\right)-2\right]=2-y\\\left(y+1\right)\left[y\left(y-1\right)-2\right]=2\left(2-z\right)\\\left(z+1\right)\left[z\left(z-1\right)-2\right]=3\left(2-x\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(x^2-x-2\right)=2-y\\\left(y+1\right)\left(y^2-y-2\right)=2\left(2-z\right)\\\left(z+1\right)\left(z^2-z-2\right)=3\left(2-x\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\left(x-2\right)=2-y\\\left(y+1\right)^2\left(y-2\right)=2\left(2-z\right)\\\left(z+1\right)^2\left(z-2\right)=3\left(2-x\right)\end{cases}}\)

Nhân các vế của 3 phương trình với nhau ta được:

\(\left(x+1\right)^2\left(x-2\right)\left(y+1\right)^2\left(y-2\right)\left(z+1\right)^2\left(z-2\right)=6\left(2-y\right)\left(2-z\right)\left(2-x\right)\)

\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=-6\left(y-2\right)\left(z-2\right)\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\left(y-2\right)\left(x-2\right)\left(z-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left[\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\right]=0\)

Vì \(\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6>0\)

Nên \(\left(x-2\right)\left(y-2\right)\left(z-2\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y-2=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\\z=2\end{cases}}}\)

Vậy x = y = z = 2

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