6/8 = 3/4 ; 39/65 = 3/5 ; 70/42 = 5/3 ; 84/231 = 4/11 ; 117117/504504 = 13/56 ; 7/35 = 1/5 ; 112/140 = 4/5 ; 165/110 = 3/2 ; 30/105 = 2/7 ; 153153/414414 = 17/46

6/8=3/4

Tử số chưa rút gọn là :

      165 : ( 7 + 4 ) x 7 = 105

Mẫu số chưa rút gọn là :

      165 - 105 = 60

Phân số chưa rút gọn là \(\frac{105}{60}\)

105/60

Tử số chưa rút gọn là :

165 : (7 + 4) x 7 = 105

Đặt \(A=1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103\)

\(=1\cdot2\cdot\left(3+1\right)+2\cdot3\cdot\left(4+1\right)+\cdots+100\cdot101\cdot\left(102+1\right)\)

\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

Đặt \(B=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\)

\(=\left(2-1\right)\cdot2\cdot\left(2+1\right)+\left(3-1\right)\cdot3\cdot\left(3+1\right)+\cdots+\left(101-1\right)\cdot101\cdot\left(101+1\right)\)

\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+101\left(101^2-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+3^3+\cdots+101^3\right)-1-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1+2+3+\cdots+101\right)\)

\(=\left(1+2+3+\cdots+101\right)^2-\left(1+2+3+\cdots+101\right)\)

\(=\left\lbrack101\cdot\frac{102}{2}\right\rbrack^2-101\cdot\frac{102}{2}=\left(101\cdot51\right)^2-101\cdot51\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+100\cdot101\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+100\left(100+1\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{100\cdot101}{2}=\frac{100\cdot101\cdot201}{6}+50\cdot101\)

\(=50\cdot101\cdot67+50\cdot101=50\cdot101\cdot68\)

Ta có: A\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

=B+C

\(=\left(101\cdot51\right)^2-101\cdot51+50\cdot101\cdot68\)

\(=101^2\cdot51^2-101\cdot51+50\cdot101\cdot68=101\left(101\cdot51^2-51+50\cdot68\right)=101\cdot266050\)

Đặt \(D=1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+101^2\left(101-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2^2+3^2+\cdots+101^2\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1^2+2^2+\cdots+101^2\right)\)
\(=\left(1+2+\cdots+101\right)^2-101\cdot\frac{\left(101+1\right)\left(2\cdot101+1\right)}{6}\)

\(=\left(101\cdot\frac{102}{2}\right)^2-101\cdot17\cdot2023=101^2\cdot51^2-101\cdot17\cdot2023\)

\(=101\cdot17\left(101\cdot17\cdot3^2-2023\right)=101\cdot17\cdot13430\)

Ta có: \(\frac{1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103}{1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2}\)

\(=\frac{101\cdot266050}{101\cdot17\cdot13430}=\frac{1565}{1343}\)

Hướng dẫn giải:

Hướng dẫn giải:

Hướng dẫn giải:

Hướng dẫn giải:

Nếu tử số là \(2\)phần thì mẫu số là \(7\)phần.

Hiệu số phần bằng nhau là: 

\(7-2=5\)(phần) 

Tử số là: 

\(30\div5\times2=12\)

Mẫu số là:

\(12+30=42\)

Phân số cần tìm là: \(\frac{12}{42}\).