Tính \(A=1+10+10^2+10^3+.....+10^{100}\)
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Tính nhanh A = (1^100+2^100+…+10^100).(5^10-25^5)
B = (2^5+3^5+4^5).(1^2+2^2+…+100^2).(4^10-2^20)
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`Answer:`
\(A=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-25^5\right)\)
\(=\left(1^{100}+2^{100}+...+10^{100}\right)[5^{10}-\left(5^2\right)^5]\)
\(=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-5^{10}\right)\)
\(=\left(1^{100}+2^{100}+...+10^{100}\right).0\)
\(=0\)
\(B=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(4^{10}-2^{20}\right)\)
\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(2^{2.10}-2^{20}\right)\)
\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right).0\)
\(=0\)
tính nhanh
A=(1^100+2^100+…+10^100).(5^10-25^5)
B=(2^5+3^5+4^5).(1^2+2^2+…+100^2).(4^10-2^20)
Tinh:
A= (10^3-1}^1.(10^3-2)^2.(10^3-3)^3.........(10^3-100)^100
bài 1: Tính
a, \(A=1+3+3^2+3^3+....3^n\)
b, \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+....+\frac{1}{10^{100}}\)
a.
\(A=1+3+3^2+3^3+...+3^n\)
\(3A=3+3^2+3^3+3^4+...+3^{n+1}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{n+1}\right)-\left(1+3+3^2+3^3+...+3^n\right)\)
\(2A=3^{n+1}-1\)
\(A=\frac{3^{n+1}-1}{2}\)
b.
\(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\)
\(10B=10+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^{98}}+\frac{1}{10^{99}}\)
\(10B-B=\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\right)-\left(10+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^{98}}+\frac{1}{10^{99}}\right)\)
\(9B=\frac{1}{10^{100}}-10\)
\(B=\frac{\frac{1}{10^{100}}-10}{9}\)
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1. TÌM X :
10+3.(x-6)=5^10:5^8
2.TÍNH
A,100-[150-8.(7-4)^2]
B,(-999)+[(-23)+999]-[10-(-67)]
Bài 1:
$10+3(x-6)=5^{10}:5^8=5^2=25$
$3(x-6)=25-10=15$
$x-6=15:3=5$
$x=5+6=11$
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Bài 2:
a. $100-[150-8(7-4)^2]=100-(150-8.3^2)=100-150+8.3^2$
$=-50+72=72-50=22$
b. $=-999-23+999-10-67=(-999+999)-10-(67+23)$
$=0-10-90=-(10+90)=-100$
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Tính :
a)
A= 1-2+3-4+5-6+...+99-100
b)
B= (10-1).(10-2).(10-3)....(10-99).(10-100)
A = 1 - 2 + 3 - 4 + 5 - 6 + ... + 99 - 100
Biểu thức A có : (100 - 1) : 1 + 1 = 100 (số hạng)
Nhóm hai số hạng thành 1 nhóm ta được : 100 : 2 = 50 (nhóm)
=> A = (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100)
=> A = (-1) + (-1) + (-1) + ... + (-1) (50 thừa số -1)
=> A = -1 . 50
=> A = -50
Vậy A = -50
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Bài 1: Tính:
a) 10 000 x 9 + 1 000 x 8 + 100 x 7 + 10 x 6 + 5
b) 100 000 x 3 + 100 x 6 + 10 x 5 + 7
c) 100 000 x 6 + 10 000
Bài 2: Tìm X:
a) X + 77=600x30
b) 8950 : X=10:2
c) 5 x X=8765
Bài 1 :
a) = 98765
b) = 300657
c) 610000
Bài 2:
a) X + 77 = 1800
X = 1800 - 77
X = 1723
b) 8950 : x = 5
x = 8950 : 5
x = 1790
c) x = 8765 : 5
x = 1753
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a,98765 b,300657 c,610000 Bài 2 : a, 17923 b, 1790 c, 1753
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Bài 1 : tính
A.98765
B.3000657
C.6100000
Bài 2 : tim X
A. 17923
B.1790
C.1753
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Xem thêm câu trả lời
tính tổng:
a) A = 1/3 + 1/3^2 + 1/3^3 +........+ 1/3^100
b) B = 10/56 + 10/140 + 10/260 +.....+ 10/1400
mọi người giúp mk vs
ai làm đc 1 trong 2 câu nhanh nhất mk sẽ tk cho
THANKS
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1-\frac{1}{3^{100}}}{2}\)
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)
\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)
\(B=\frac{15}{14}:3=\frac{5}{14}\)
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a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\frac{3}{14}\)
\(\Rightarrow B=\frac{5}{14}\)
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\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=1-\frac{1}{3^{100}}\)\(\Rightarrow2A=\left(1-\frac{1}{3^{100}}\right)\Rightarrow A=\frac{1}{2}\times\left(1-\frac{1}{3^{100}}\right)\)
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A=(1^100+2^100+…+10^100).(5^10-25^5)
B=(2^5+3^5+4^5).(1^2+2^2+…+100^2).(4^10-2^20)