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

Đặt \(A=\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)

\(A=\frac{\left(3\cdot7\cdot13\cdot37\right)\cdot39-10101\cdot1}{50\cdot10101+7\cdot10101}\)

\(A=\frac{10101\cdot39-10101\cdot1}{10101\cdot\left(50+7\right)}\)

\(A=\frac{10101\cdot\left(39-1\right)}{10101\cdot57}\)

\(A=\frac{10101\cdot38}{10101\cdot57}\)

\(A=\frac{38}{57}=\frac{38:19}{57:19}=\frac{2}{3}\)

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

\(131313-10101-20202=13\cdot10101-10101-2\cdot10101=10101\left(13-1-2\right)=10101\cdot10=101010\)