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

 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

Dương Lam Hàng Bạn cất công thật, cảm ơn nhé

a, Ta có: \(\left(\sqrt{2}+\sqrt{3}\right)^2\)= \(2+2\sqrt{6}+3=5+2\sqrt{6}\)

Lại có \(3^2=9=5+4\)mà \(2\sqrt{6}>4\)

suy ra \(\left(\sqrt{2}+\sqrt{3}\right)^2>9\)

suy ra \(\sqrt{2}+\sqrt{3}>3\)

b, Ta có: \(\left(\sqrt{11}-\sqrt{3}\right)^2=11-2\sqrt{33}+3=14-2\sqrt{33}\)

Lại có: \(2^2=4=14-10\)mà \(2\sqrt{33}>10\)

suy ra \(\left(\sqrt{11}-\sqrt{3}\right)^2< 2^2\)

suy ra \(\sqrt{11}-\sqrt{3}< 2\)

#)Giải :

a) √2 +√3 = √( √2 + √3 )² = √( 5 + 2√6 ) = √( 5 + √24 ) 

3 = √9 = √( 5 + √16 )                                          

=> √2 + √3 > 3

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

suy ra \(x^2+x+1=\left(x+1\right)^2\)

suy ra \(x^2+x+1\)= \(x^2+2x+1\)

suy ra \(x=2x\)

suy ra \(2x-x=0\)

suy ra \(x=0\)

\(\sqrt{\frac{x+1}{x-1}}\)\(đkxđ\Leftrightarrow\hept{\begin{cases}\frac{x+1}{x-1}\ge0\\x-1\ne0\end{cases}}\)

\(\frac{x+1}{x-1}\ge0\)\(\Rightarrow\orbr{\begin{cases}x+1\ge0;x-1\ge0\\x+1< 0;x-1< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge-1;x\ge1\\x< -1;x< 1\end{cases}\Rightarrow\orbr{\begin{cases}x\ge1\\x< -1\end{cases}}}\)

Và \(x-1\ne0\Rightarrow x\ne1\)

\(\Rightarrow x>1\)Hoặc  \(x< -1\)

Nối BM
Xét tam giác BMD vuông tại D, có: BD² = BM² - MD² (1)
Xét tam giác MCD vuông tại D, có: DC² = MC²- MD² (2)
Từ (1) và (2) => BD² - DC² = BM²- MD² - MC² + MD² = BM² - MC² = BM² - AM² (vì AM=CM) = AB²

=> AB² = BD²- DC² (đpcm)

Ta có bđt quen thuộc sau \(\frac{x}{y+z}< \frac{x+m}{y+z+m}\) 

Áp dụng ta được \(\frac{a}{b+c}< \frac{a+a}{a+b+c}=\frac{2a}{a+b+c}\)
Chứng minh tương tự \(\frac{b}{c+a}< \frac{2b}{a+b+c}\)

                                     \(\frac{c}{a+b}< \frac{2c}{a+b+c}\)

Do đó \(VT< \frac{2a+2b+2c}{a+b+c}=2\)

Ta đi chứng minh VP > 2 

Áp dụng bđt Cô-si có \(a+\left(b+c\right)\ge2\sqrt{a\left(b+c\right)}\)

                             \(\Rightarrow\sqrt{a\left(b+c\right)}\le\frac{a+b+c}{2}\)

                             \(\Rightarrow\sqrt{\frac{b+c}{a}}\le\frac{a+b+c}{2a}\)

                             \(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)

Chứng minh tương tự \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\)

                                    \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

Cộng 3 vế lại ta được \(VP\ge\frac{2a+2b+2c}{a+b+c}=2\)

Do đó \(VP\ge2>VT\)

\(\Rightarrow VT< VP\left(Q.E.D\right)\)

Dấu "=" không xảy ra