\(\frac{327\cdot412+400}{328\cdot412-12}\)

\(=\frac{327\cdot412+400}{\left(327+1\right)\cdot412-12}\)

\(=\frac{327\cdot412+400}{327\cdot412+\left(412-12\right)}\)

\(=\frac{327\cdot412+400}{327\cdot412+400}\)

\(=1\)

  327.412+400 / 328.412-12

= 327.412+400 / (327+1).412-12

= 327.412+400 / 327.412+1.412-12

= 327.412+400 / 327.412+412-12

= 400/412-12

= 400/400

= 1

a)=15.2154/1505

=6462/301

b)=461179719/138076

c,=134724+400/135136-12

=135124/135124

=1

a)=15.(21/43+33?55)=15.234/215=16 14/43
b)=328.412-412+400/328.412-12=328.412-12/328.412-12=1
c)7217721 hay là 721721 vậy?
 

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

a) Ta có: \(A=\dfrac{37^3+12^3}{49}-37\cdot12\)

\(=\dfrac{\left(37+12\right)\left(37^2-37\cdot12+12^2\right)}{49}-37\cdot12\)

\(=37^2-2\cdot37\cdot12+12^2\)

\(=\left(37-12\right)^2\)

\(=25^2=625\)

\(a,=2^3\left(17-14\right)=2^3\cdot3=24\\ b,=80-\left(130-8^2\right)=80-\left(130-64\right)=80-66=14\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{11\cdot12}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)

\(=1-\frac{1}{12}=\frac{11}{12}\)

Vậy \(A=\frac{11}{12}\)

Chúc bạn học tốt ^^!!!

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)

\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{11\times12}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=1-\frac{1}{12}=\frac{11}{12}\)

Chúc bạn học tốt!

#Huyền#

 A=\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+.....+\(\frac{1}{132}\)

A= \(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+.....+\(\frac{1}{11x12}\)

A= 1  -\(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{11}\)-\(\frac{1}{12}\)

A= 1-\(\frac{1}{12}\)

A=\(\frac{11}{12}\)