bó con mẹ nó giò rồi

Khó quá!Xin lỗi bạn.

\(\sqrt{2}^{2018}+2.\sqrt{2^{1009}.3^{1009}}+\sqrt{3}^{2018}+\sqrt{2}^{2018}-2\sqrt{2^{1009}.3^{1009}}+\sqrt{3}^{2018}\)

\(2\sqrt{2}^{2018}+2\sqrt{3}^{2018}=A\)

\(\sqrt{2^{2018}}=2^{1009}\)

\(\sqrt{3^{2018}}=3^{1009}\)

\(A=2.2^{1009}+2.3^{1009}\)

Đặt:   \(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}\)

Ta có:   \(\frac{1}{\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)       với \(\forall k\inℕ^∗\)

Do đó ta có:    \(A>2\left[\left(\sqrt{2019}-\sqrt{2018}\right)+\left(\sqrt{2018}-\sqrt{2017}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)\right]+1\)

  \(=2\left(\sqrt{2019}-\sqrt{2}\right)+1=2\sqrt{2019}-2\sqrt{2}+1>2\sqrt{2019}-3+1>2\sqrt{2019}-2\)

   \(>2\sqrt{2018}-2=2\left(\sqrt{2018}-1\right)\)

=>  đpcm

Bài 2 xét x=0 => A =0

xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)

để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)

=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?

1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)

=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)

\(a+2=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)

\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)

=> \(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)}+...\)

=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)

=> M=0

Vậy M=0 

Ta có \(\sqrt{x}+\sqrt{y-2}=2\)=> \(\left(\sqrt{x}-1\right)+\left(\sqrt{y-2}-1\right)=0\)

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

=>Tương tự với các PT còn lại

\(\frac{y-3}{\sqrt{y+1}+2}+\frac{z-4}{\sqrt{z-3}+1}=0\left(2\right)\)

\(\frac{z-4}{\sqrt{z+5}+3}+\frac{x-1}{\sqrt{x+3}+2}=0\left(3\right)\)

Ta thấy \(x=1;y=3;z=4\)là nghiệm của 3 PT

Với \(x\ne1;y\ne3;z\ne4\)

Theo nguyên lí diricle ta luôn có : 

trong 3 số x-1;y-3;z-4 luôn có 2 số cùng dấu 

=> 2 trong 3 PT trên vô nghiệm 

Vậy x=1;y=3;z=4

1.

đk: \(x\ge2\)

Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:

ta có : \(x^3-3x^2+2y^3-6x=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)

ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)

\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)

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

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

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

\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(2\right)\\2\sqrt{x+2}=-x\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)

\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(x=y\Rightarrow M=x^{11}-x^{2018}\)

\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)