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Ta có : A = 1 + 2 + 3 + ... + 2008
\(A=\frac{\left(2008+1\right)\left[\left(2008-1\right)\div1+1\right]}{2}\)
\(A=\frac{2009.2008}{2}\)
\(A=2017036\)
Ta có: B = 1 + 2 + 3 + ... + 1010
\(B=\frac{\left(1010+1\right)\left[\left(1010-1\right):1+1\right]}{2}\)
\(B=\frac{1011.1010}{2}\)
\(B=510555\)
\(A=1+2+3+4+5+...+2008\)
\(A=\left(2008+1\right)\left(\left(2008-1\right):1+1\right):2=2009.2008:2\)
\(=2009.1004=2017036\)
\(B=1+2+3+4+...+1010\)
\(B=\left(1010+1\right)\left(\left(1010-1\right):1+1\right):2=1011.\left(1010:2\right)\)
\(=1011.505=510555\)
\(C=2+5+8+11+...+302\)
\(C=\left(302+2\right)\left(\left(302-2\right):3+1\right):2=304.101:2\)
\(=15352\)
\(D=3+3^2+3^3+3^4+...+3^{2019}\)
\(3D=3^2+3^3+3^4+...+3^{2020}\)
\(3D-D=\left(3^2+3^3+3^4+...+3^{2020}\right)-\left(3+3^2+3^3+3^4+...+3^{2019}\right)\)
\(2D=3^{2020}-3\)
\(\Rightarrow D=\frac{3^{2020}-3}{2}\)
\(E=4^{10}+4^{11}+4^{12}+...+4^{100}\)
\(4E=4^{11}+4^{12}+4^{13}+...+4^{101}\)
\(4E-E=\left(4^{11}+4^{12}+4^{13}+...+4^{101}\right)-\left(4^{10}+4^{11}+4^{12}+...+4^{100}\right)\)
\(3E=4^{101}-4^{10}\)
\(E=\frac{4^{101}-4^{10}}{3}\)
Từ sau bn ra ít thôi nha
Câu 1:
a) 52.49+48.38-52.20-48.11
=52.49-52.20+48.38-48.11
=52.(49-20)+28.(38-11)
=52.29+28.27
=2264
b)2.54.12+4.6.68-3.8.22
=24.(54+68-22)
=24.100
=2400
c)24+31.24.64+24.8
=24.(31+64+8)
=24.103
=1648
d)19:3+28:3+13:3
=(19+28+13):3
=60:3
=20
e)32.15-{[(102.15-15.90)-43.2)
=32.15-{[15.(102-90)-43.2)]}
=32.15-{[150-128]}
=32.15-22
=135-22
=113
a)32<2x<128
25<2x<27
Vì 25<26<27
Suy ra:2x=26
Vậy x=6
b)(x+1)2.3245:43-20(ko có nghĩa)
c)92-3(x-3)=157:156
92-3(x-3)=15
3(x-3)=66
x-3=22
x=25
Vậy x=25
d)3x.310=317
3x=317:310
3x=37
Vậy x=7
e)81.3x+1=315
\(3^{x+1}=\frac{35}{9}\)
\(3^{x+1}=3^{1,23621727}\)(cái này ko đổi ra được phân số)
x+1=1,23621727
Rồi tự tính nha
Chơi câu khó nhất
D = 4 + 42 + 43 + ... + 4n
4D = 42 + 43 + ... + 4n+1
3D = 4n+1 - 4
D = \(\frac{4^{n+1}-4}{3}\)
a) \(A=2+2^2+2^3+....+2^{100}\)
\(2A=2^2+2^3+2^4+....+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+....+2^{101}\right)-\left(2+2^2+....+2^{100}\right)\)
\(A=2^{101}-2\)
B) \(B=1+3+3^2+3^3+...+3^{2009}\)
\(3B=3+3^2+3^3+3^4+...+3^{2010}\)
\(3B-B=\left(3+3^2+3^3+3^4+...+3^{2010}\right)-\left(1+3+3^2+...+3^{2009}\right)\)
\(2B=3^{2010}-1\)
\(B=\frac{3^{2010}-1}{2}\)
C) \(C=1+5+5^2+....+5^{1998}\)
\(5C=5+5^2+5^3+...+5^{1999}\)
\(5C-C=\left(5+5^2+5^3+...+5^{1999}\right)-\left(1+5+5^2+...+5^{1998}\right)\)
\(4C=5^{1999}-1\)
\(C=\frac{5^{1999}-1}{4}\)
D) \(D=4+4^2+4^3+...+4^n\)
\(4D=4^2+4^3+4^4+...+4^{n+1}\)
\(4D-D=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3D=-4\)
\(D=\frac{-4}{3}\)
Ý D mk ko bít đúng ko
hok tốt k mk nhé
\(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+...+2^{101}\)
\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(A=2^{101}-2\)'
\(B=1+3+3^2+3^3+...+3^{2009}\)
\(3B=3+3^2+3^3+...+3^{2010}\)
\(3B-B=3^{2010}-1\)
\(2B=3^{2010}-1\)
\(B=\frac{3^{2010}-1}{2}\)
\(C=1+5+5^2+5^3...+5^{1998}\)
\(5C=5+5^2+...+5^{1999}\)
\(5C-C=5^{1999}-1\)
\(4A=5^{1999}-1\)
\(A=\frac{5^{1999}-1}{4}\)
\(D=4+4^2+4^3+...+4^n\)
\(4D=4^2+4^3+...+4^{n+1}\)
\(4D-D=4^{n+1}-4\)
\(3D=4^{n+1}-4\)
\(D=\frac{4^{n+1}-4}{3}\)
\(C=1+3+3^2+3^3+...+3^{15}\)
=> \(3C=3+3^2+3^3+3^4+...+3^{16}\)
=> \(3C-C=3^{16}-1\)
=>\(2C=3^{16}-1\)
=>\(C=\frac{3^{16}-1}{2}\)
Câu D tương tự