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a) 5x . ( 53) 2 = 625
5x . 56 = 625
mà 625 = 54
Suy ra : x + 6 = 4
x = 4 - 6
x = -2
b) (-3/4 )3x - 1 = 256/81
(-3/4 )3X - 1 = (-3/4)-4
SUY RA : 3X - 1 = -4
3X = -4 + 1 = -3
X = -3 : 3
X = -1
C ) (8x - 1 )2n+1 = 52n+1
SUY RA : 8X - 1 = 5
8X = 5 + 1
8 X = 6
X = 6 : 8
X = 3/4
d) (x - 2/9 )2 = 4/9
mà 4/9 = 2/32
SUY RA : x - 2/9 = 2/3
x = 2/3 + 2/9
x = 24/27
Câu e mình không bít làm bn chịu khó suy nghĩ nha !
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(5A=5^1+5^2+5^3+5^4+...+5^{51}\)
\(4A=5A-A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
b/
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{100}\)
\(\frac{1}{2}B=B-\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\)
\(B=\frac{1}{2}B\cdot2=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
\(B=1-\frac{1}{2^{99}}\)
A=2100-299+298-297+.....+22-2
=>2A=2101-2100+299-298+.....+23-22
=>2A+A=2101-2100+299-298+.....+23-22+2100-299+298-297+....+22-2
=>3A=2201-2
=>A=\(\frac{2^{201}-2}{3}\)
B=3100-399+398-397+....+32-3+1
=>3B=3101-3100+399-398+...+33-32+3
=>3B+B=3101-3100+399-398+...+33-32+3+3100-399+398-397+....+32-3+1
=>4B=3101+1
=>B=\(\frac{3^{101}+1}{4}\)
Câu a : Đặt 2A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2
=> 2A - A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - ( 2^100 - 2^99 + 2^98 - 2^97 +...+ 2^2 - 2)
=> A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - 2^100 + 2^99 - 2^98 + 2^97 -...- 2^2 + 2
=> A= = 2^101 -2(2^100 + 2^98 + 2^96 +...+ 2^2) + 2(2^99 + 2^97 + 2^95 +...+ 2^3) +2
Câu b : Làm tương tự như trên
BẤM ĐÚNG CHO MÌNH NHA
M = \(2^{100-}2^{99}+2^{98}-...+2^2-2\)
\(2M=2^{101}-2^{100}+2^{99}-2^{98}+2^{97}+...+2^3-2^2\)
\(2M+M=2^{101}-2\)
\(M=\frac{2^{101}-2}{3}\)
N=\(3^{100}-3^{^{ }99}+3^{98}-3^{97}+...+3^2-3+1\)
\(3N=3^{101}-3^{100}+3^{99}-3^{98}+3^{97}+...+3^3-3^2+3\)
3N+N= 4N = \(3^{101}+1\)
N=\(\frac{3^{101}+1}{4}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3B-B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(3A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
Chúc bạn học tốt ~