Bài 3. Tính nhanh :
( 12+22+33+........+20202) . ( 91 - 273 : 3 )
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\(\left(1^2+2^2+3^2+...+1000000^2\right).\left(91-273:3\right)\\ =\left(1^2+2^2+3^2+...+1000000^2\right).0=0\)
(12+22+32+..+10000002).(91-273:3)
=(12+22+32+..+10000002).(91-91)
=(12+22+32+..+10000002).0
=0
1.
$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$
2.
$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$
3. Không phù hợp để tính nhanh
4.
$=15^8-(15^8-1)=1$
5.
$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$
$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$
$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$
6:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)
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.
$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$
$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$
$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$
$=2020+2019+2018+2017+...+2+1$
$=\frac{2020.2021}{2}=2041210$
91-273:3=91-91=0
=>(12+22+32+...+1002)(91-273:3)=(12+22+32+...+1002).0=0
a: \(13\cdot65+35\cdot12\)
\(=13\cdot65+35\cdot13-35\)
=1300-35
=1265
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{4}\)
=(1/3+2/3)+(1/4+1/4)+(1/5+4/5)
=1+1+1/2
=5/2
2*32*12*4*6*41= ( 2 * 12) * (32*41) * (4*6)
= 24 * 1312 * 24
= 24 *24 * 1312
= 576 * 1312
= 755712
2*33*7+7*2*45+7*22*2 = 7 * ( 2*33+ 2*45 + 22*2)
= 7 * ( 66+90+44)
= 7* 200 = 1400
17*85+15*17-120 = 17 * ( 85 + 15) - 120
= 17*100- 120
=1700-120
= 1580
(12+22...+10002).(91-273:3)
=(12+22...+10002).(91-91)
=(12+22...+10002).0
=0
91 -273:3=91-91=0