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Đặt \(A=1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\)
\(\frac{1}{3}A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+.....+\frac{1}{3}^{101}\)
\(\frac{1}{3}A-A=\left(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}\right)-\left(1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\right)\)
\(\frac{1}{3}A-A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}-1-\frac{1}{3}-\frac{1}{3}^2-\frac{1}{3}^3-....-\frac{1}{3}^{100}\)
\(\frac{\left(-2\right)}{3}A=\frac{1}{3}-1\)
\(\frac{\left(-2\right)}{3}A=\frac{\left(-2\right)}{3}\Rightarrow A=1\)
Vậy ......
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ta có A=1+1/3+1/3^2+...+1/3^100
A.3= 3+1+1/3+1/3^2+...+1/3^99
A.3-A= (3+1+1/3+1/3^2+...+1/3^99)-(1+1/3+1/3^2+...+1/3^100)
A.2=3-1/3^100
A=(3-1/3^100):2
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Nếu a+3 là dương
A=3a-3-2.(a+3)+9
A=3a-3-2a+6+9
A=a+12
Nếu a+3 là âm
A=3a-3-2.(-a-3)+9
A=3a-3-(-2).a-6+9
A=5.a+9-6-3
A=5.a
T..i..c..k nha
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1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)
\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)
\(\Rightarrow3S-S=3^{2021}-1\)
\(\Rightarrow2S=3^{2021}-1\)
\(\Rightarrow S=\frac{3^{2021}-1}{2}\)
2. \(\left(3x-2\right)^3=64\)
\(\Leftrightarrow\left(3x-2\right)^3=4^3\)
\(\Leftrightarrow3x-2=4\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
[3x-2]^3=64
Ta có:64=4^3
Suy ra:3x-2=4
3x =4+2
3x=6
x=6:3
x=2
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a)
A= (-m+n-p)-(-m-n-p)
A= -m+n-p+m+n+p
A= (-m+m) +(n+n) + (-p+p)
A= 0+2n+0
A = 2n
Bài 1:
A = (-m + n - p) - (-m - n - p)
A = -m + n - p + m + n + p
A = (-m + m) + (n + n) - (p - p)
A = 2n
Với n = -1 => A = 2(-1) = -2
Bài 2:
A = (-2a + 3b - 4c) - (-2a -3b - 4c)
A = -2a + 3b - 4c + 2a + 3b + 4c
A = (-2a + 2a) + (3b + 3b) - (4c - 4c)
A = 6b
Với b = -1 => A = 6(-1) = -6
Bài 3:
a) A = (a + b) - (a - b) + (a - c) - (a + c)
A= a + b - a + b + a - c - a - c
A = (a - a + a - a) + (b + b) - (c + c)
A = 2(b - c)
b) B = (a + b - c) + (a - b + c) - (b + c - a) - (a - b - c)
B = a + b - c + a - b + c - b - c + a - a + b + c
B = (a + a + a - a) + (b - b - b + b) - (c - c + c - c)
B = 2a
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\(A=1+3+3^2+3^3+...+3^{20}\)
=> \(3A=3+3^2+3^3+3^4+...+3^{21}\)
=> \(3A-A=3^{21}-1\)
=> \(2A=3^{21}-1\)
=> \(A=\frac{3^{21}-1}{2}\)