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a, Ta có :
A = 1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100
2A = 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101
A = 2A – A = ( 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 ) –( 1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 )
= 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 – 1 - 2 - 2 2 - 2 3 - 2 4 - . . . - 2 99 - 2 100
= 2 101 - 1
Vậy A = 2 101 - 1
b, Ta có.
B = 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99
5 2 B = 5 2 ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )
25B = 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101
25B – B = ( 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 ) – ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )
24B = 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 – 5 - 5 3 - 5 5 - . . . - 5 97 - 5 99
24B = 5 101 - 5
B = 5 101 - 5 24 = 5 5 100 - 1 24
Vậy B = 5 5 100 - 1 24
Bài làm
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}+2=\frac{x+50}{23}+\frac{x+49}{24}+2\)
\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)=\left(\frac{x+50}{23}+1\right)+\left(\frac{x+49}{24}+1\right)\)
\(\Leftrightarrow\left(\frac{x+52}{21}+\frac{21}{21}\right)+\left(\frac{x+51}{22}+\frac{22}{22}\right)=\left(\frac{x+50}{23}+\frac{23}{23}\right)+\left(\frac{x+49}{24}+\frac{24}{24}\right)\)
\(\Leftrightarrow\frac{x+52+21}{21}+\frac{x+51+22}{22}=\frac{x+50+23}{23}+\frac{x+49+24}{24}\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}=\frac{x+73}{23}+\frac{x+73}{24}\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)
\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)
\(\Leftrightarrow x+73=0\)
\(\Leftrightarrow x=-73\)
Vậy x = -73
# Học tốt #
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}-\frac{x+50}{23}-\frac{x+49}{24}=0\)
\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)-\left(\frac{x+50}{23}+1\right)-\left(\frac{x+49}{24}+1\right)=0\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)
\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)
=> \(x+73=0\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\ne0\right)\)
<=> x=-73
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
Bài 2:
a. $=58(75+50-25)=58.100=5800$
b. $=27.8+4.9-5=216+36-5=247$
Bài 3:
a.
$(x-38):16=12$
$x-38=16\times 12=192$
$x=192+38 = 230$
b.
$10+2x=4^5:4^3=4^2=16$
$2x=16-10=6$
$x=6:2=3$
c.
$440+2(125-x)=546$
$2(125-x)=546-440=106$
$125-x=106:2=53$
$x=125-53=72$
d.
$(x-15):5+20=22$
$(x-15):5=22-20=2$
$x-15=2\times 5=10$
$x=10+15=25$
e.
$2x-138=2^3.3^2=8.9=72$
$2x=72+138 = 210$
$x=210:2=105$
a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
a/B=5^101-5
b/A=2^6-2^2