Quy đồng và cho nó vô ngoặc cho dễ ha!

= 1/21 +( 2/36 + 1/36) + ...+ 1/120

= 1/21 + 3/36 + ....+ 1/120

= 99/756 + ... + 1/120

=

tới đây thì hơi bí rồi! Tự làm nha! Chỉ gợi ý cho bạn thôi!

Thanks

Quy đồng và cho nó vô ngoặc cho dễ ha!

= 1/21 +( 2/36 + 1/36) + ...+ 1/120

= 1/21 + 3/36 + ....+ 1/120

= 99/756 + ... + 1/120

=

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/99 + 1/101)

= 1/2 . (1/1 - 1/101)

= 1/2 . 100/101

= 50/101

oh

1. Lan can be helped by Nam

2. Lan and Hoa will be helped by Nam

3. We should be helped by Nam

1. Lan can be helped

2. Lan and Hoa will be helped

3. We should be helped

1. We must be helped

2. We used to be helped 

3. Lan has to helped 

1.      Nam must help us.                                                                    →………We must be helped…………………………………………

2.      Nam used to help us.                                                                         →……We used to be helped ……………………………………………

3.      Nam has to help Lan.                                                         →………… Lan has to helped………………………………………

Ta có : ( 2x - 1 )²⁰²⁰ = ( 2x - 1 )²⁰²¹

=> ( 2x - 1 )²⁰²¹ - ( 2x - 1 )²⁰²⁰ = 0

=> ( 2x - 1 )²⁰²⁰ . [( 2x -1 )¹ - 1 ] = 0

=>  2x - 1 = 0               2x = 1                     x = 1/2

 hoặc                  =>                         =>

     2x - 1 = 1                2x = 2                     x =1

Vậy x = 1 hoặc x = 1/2

\(\frac{1}{a}-1=\frac{a+b+c}{a}-\frac{a}{a}=\frac{b+c}{a}\)

Tương tự : \(\frac{1}{b}-1=\frac{c+a}{b};\frac{1}{c}-1=\frac{a+b}{c}\)

Nhân theo vế ta đc :

\(VT=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Áp dụng bđt Cauchy :

\(VT\ge\frac{8abc}{abc}=8\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

theo bài ra ta có:

\(\dfrac{6}{x+1}.\dfrac{x-1}{3}=\dfrac{6x-6}{3x+1}\\ =\dfrac{6x+2-8}{3x+1}\\ =\dfrac{2\left(3x+1\right)-8}{3x+1}\\ =2-\dfrac{8}{3x+1}\)

để \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\) là số nguyên

=> \(\dfrac{8}{3x+1}\) nguyên

\(8⋮3x+1\\ \Rightarrow3x+1\inƯ_{\left(8\right)}=\left\{-1;1;2;-2;4;-4;8;-8\right\}\)

ta có bảng sau:

mà x là số nguyên

=> x ={0;-1;1;-3}

vậy x ={0;1;-1;-3}

\(\left(x+1\right)^3+\left(x-1\right)^3=\left(x-1\right)\left(x+1\right)+4\)

\(\Rightarrow\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x+1\right)\left(x-1\right)-4=0\)

\(\Rightarrow2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)-x^2+1-4=0\)

\(\Rightarrow2x\left(x^2+3\right)-x^2+1-4=0\)

\(\Rightarrow2x^3+6x-x^2-3=0\)

\(\Rightarrow\left(2x^3+6x\right)-\left(x^2+3\right)=0\)

\(\Rightarrow2x\left(x^2+3\right)-\left(x^2+3\right)=0\)

\(\Rightarrow\left(x^2+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=-3\left(L\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)

5 nha ban k cho mk nha

1+1+1+1+1=5

1 + 1 + 1 + 1 + 1

= 1 x 5

= 5