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! :))

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

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

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

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

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

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

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

\(\Leftrightarrow x+1=4034\)

\(\Leftrightarrow x=4033\)

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