Ta co:

1/2+1/6+1/12+1/20+1/30+1/42

=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

Ta co: 6/7=6.10101/7.10101=60606/70707 

Vi 60606/70707=60606/70707 nen 6/7=60606/70707

Vay 1/2+1/6+1/12+1/20+1/30+1/42 = 60606/70707

ta có:

\(\frac{60606}{70707}=\frac{6}{7}\)

ta có:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}=\)

\(1-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{7}=\frac{6}{7}\)

suy ra 6/7=6/7

suy ra \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{60606}{70707}\)

\(\left(\frac{151515}{60606}+\frac{151515}{121212}+\frac{151515}{202020}+\frac{151515}{303030}+\frac{151515}{242424}\right)\cdot\frac{28}{12}\)

\(=\left(\frac{5}{2}+\frac{5}{4}+\frac{3}{4}+\frac{1}{2}+\frac{5}{8}\right)\cdot\frac{28}{12}\)

\(=\left[\left(\frac{5}{2}+\frac{1}{2}\right)+\left(\frac{5}{4}+\frac{3}{4}\right)+\frac{5}{8}\right]\cdot\frac{28}{12}\)

\(=\left(3+2+\frac{5}{8}\right)\cdot\frac{28}{12}\)

\(=\left(5+\frac{5}{8}\right)\cdot\frac{28}{12}\)

\(=\frac{45}{8}\cdot\frac{28}{12}\)

\(=\frac{105}{8}\)

so sánh hả bạn?

ta có : 191919 + 20202 = 212121

           353535 + 70707 = 424242

vì 212121 < 424242

=> 191919 + 20202 < 353535 + 70707

Mình cần bài toán rút gọn phân số chứ ko phải so sánh

x0x0x : X x 3 – 20202

= 10101 x 3 – 20202

= 30303 – 20202

= 10101

a. Ta có: \(17^2-14.17+49=17^2-2.7.17+7^2=\left(17-7\right)^2=10^2=100\)

b. \(2021^2-2020^2=\left(2021-2020\right)\left(2021+2020\right)=4041\)

a) \(153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)

b)

\(\left(2020^2-2019^2\right)+\left(2018^2-2017^2\right)+...+\left(2^2-1^2\right)\\ =\left(2020+2019\right)\left(2020-2019\right)+\left(2018+2017\right)\left(2018-2017\right)+...+\left(2+1\right)\left(2-1\right)\\ =2020+2019+2018+2017+...+2+1\\ =\dfrac{\left(2020+1\right)2020}{2}=2041210\)

Lời giải:

a. $153^2-53^2=(153-53)(153+53)=100.206=20600$

b. 

$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$

$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$

$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$

$=2020+2019+2018+2017+...+2+1$

$=\frac{2020.2021}{2}=2041210$

a) 153²-53²=(153-53)(153+53)=100.206=20600

C

C

C

a) 544544 – 444444 = 544.1001 – 444.1001 = 1001.(544 – 444) = 1001.100 = 100100

b) 131313 – 10101 – 20202 = 10101.13 – 10101 – 10101.2

= 10101.(13 – 1 – 2) = 10101.10 = 101010