\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2021}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{2021}-\sqrt{2020}}{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{2021}-\sqrt{2020}}{2021-2020}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2021}-\sqrt{2020}\)

\(=\sqrt{2021}-1\)

Gọi thời vòi 1 vòi 2 chảy đầy bể lần lượt là a ; b ( a ; b > 0 ) 

\(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{6}\\\dfrac{2}{a}+\dfrac{3}{b}=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{10}\\\dfrac{1}{b}=\dfrac{1}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=10\\b=15\end{matrix}\right.\left(tm\right)\)

đặt \(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}+\frac{1}{102}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{102}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}+\frac{1}{102}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{51}\right)\)

\(A=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)

Khó quá taa=))

Hmm

ừm đúng rồi vậy sửa lại nha

A=100-1/100

A=99,99

Với mọi \(n\in N\)ta có:

\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}\)

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

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

Áp dung vào biểu thức A ta được:

\(A=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+....+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+1+1+...+1\right)+\left(-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)(Có 99 số 1)

\(A=99-\frac{1}{100}=\frac{9899}{100}\)

1) Phương trình đó có vô số nghiệm khi \(\hept{\begin{cases}m^2-1=0\\m+1=0\end{cases}}\Leftrightarrow m=-1\)

\(\Rightarrow\)Chọn A

2) Phương trình đó có nghiệm duy nhất khi \(m^2-1\ne0\Leftrightarrow m\ne\pm1\)

\(\Rightarrow\)Chọn D.

Đặt:    \(a=x\); \(b=x-1\)

Khi đó phương trình đã cho có dạng:

              \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow\)\(a^3+b^3=a^3+b^3+3ab.\left(a+b\right)\)

\(\Leftrightarrow\)\(3ab.\left(a+b\right)=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a=0\\b=0\\a+b=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x-1=0\\x+x-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=1\\x=\frac{1}{2}\end{cases}}\left(TM\right)}\)

Kết luận:....