\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2016}{2017}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1008}{2007}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4034}\)

\(\Leftrightarrow x+1=4034\)

\(\Leftrightarrow x=4033\)

Vậy x = 4033

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)

=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2016}{2017}\right)\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2016}{2017}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{1017}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2016}{2017}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1008}{2017}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1008}{2017}\)

=> \(\frac{1}{x+1}=\frac{1}{4034}\)

Vì 1 = 1

=> x + 1 = 4034

=> x       = 4034 - 1

=> x       = 4033

Lưu ý : Dấu "." là dấu nhân

\(9xy-6x+3y=6\)

\(\Leftrightarrow3x.\left(3y-2\right)+3y=6\)

\(\Leftrightarrow3x.\left(3y-2\right)+3y-2=6-2\)

\(\Leftrightarrow3x.\left(3y-2\right)+\left(3y-2\right)=4\)

\(\Leftrightarrow\left(3y-2\right)+\left(3x+1\right)=6\)

Mà \(x,y\in Z\Rightarrow3y-2;3x+1\in Z\)

Lập bảng làm nốt

Nhầm dòng thứ 5 sửa số 6 thành số 4 cho anh 

Ta có :\(\frac{3x}{x-2}+\frac{2x+5}{\left(x-2\right)\left(x-5\right)}=\frac{x}{x-5}\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}+\frac{2x+5}{\left(x-2\right)\left(x-5\right)}=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow\frac{3x^2-15+2x+5}{\left(x-2\right)\left(x-5\right)}=\frac{x^2-2x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x^2+2x-10=x^2-2x\)\(\Leftrightarrow2x^2+4x=10\Leftrightarrow2x\left(x+2\right)=10\Leftrightarrow x\left(x+2\right)=5\)

Sau đó tự giải tiếp x.

\(ĐKXĐ:x\ne2;x\ne5\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}+\frac{2x+5}{\left(x-2\right)\left(x+5\right)}=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

\(\Rightarrow3x\left(x-5\right)+\left(2x+5\right)=x\left(x-2\right)\)

\(\Leftrightarrow3x^2-15x+2x+5=x^2-2x\)

\(\Leftrightarrow3x^2-15x+2x+5-x^2+2x=0\)

\(\Leftrightarrow2x^2-11x+5=0\)

\(\Leftrightarrow2x^2-10x-x+5\)

\(\Leftrightarrow\left(2x^2-10x\right)-\left(x-5\right)\)

\(\Leftrightarrow2x\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(2x-1\right)=0\)

Hoặc \(x-5=0\Leftrightarrow x=5\left(L\right)\)

Hoặc \(2x-1=0\Leftrightarrow x=\frac{1}{2}\left(N\right)\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)

Voi 3 chay 2 gio day be.

Ta có : 1 giờ 20 phút = 4/3 giờ 

=> 1 giờ 3 vòi chảy được :

1 : 4/3 = 3/4 bể

1 giờ vòi thứ nhất chảy được : 

1 : 6 = 1/6 bể

1 giờ vòi thứ hai chảy được :

1 : 4 = 1/4 bể

1 giờ vòi thứ ba chảy được : 

3/4 - 1/6 - 1/4 = 1/3 bể 

=> Thời gian để vòi thứ 3 chảy đầy bể là 

1 : 1/3 = 3 (giờ)

            Đáp số 3 giờ

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

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

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

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

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

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

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