Rút gọn:
A=1-2-22-23-...-2100
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Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2
Đặt A = \(1+2+2^2+2^3+2^4+....+2^{100}\)
2A = \(2\left(1+2+2^2+2^3+2^4+....+2^{100}\right)\)
= \(2+2^2+2^3+2^4+2^5+...+2^{101}\)
2A - A = \(\left(2+2^2+2^3+2^4+2^5+....+2^{101}\right)-\left(1+2^2+2^3+2^4+...+2^{100}\right)\)
= \(2^{101}-1\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)
Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B
\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)
Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C
\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)
\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)
\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)
Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
\(A=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
\(=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+\sqrt{1^2}}-\sqrt{\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}+2^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\)
\(=-2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-1-3\sqrt{2}-2\)
\(=-3\)
\(A=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2=-3\)
A = 1 - (2+22+23+...+2100)
Đặt B = 2+22+23+...+2100 => A = 1 - B
2B = 2(2+22+23+...+2100) = 22+23+...+2100+2101
=> B = 2B - B = (22+23+...+2100+2101) - (2+22+23+...+2100) = 2101 - 2
=> A = 1 - (2101 - 2) = 3 - 2101
Ta có: \(2A=2-2^2-2^3-...-2^{101}\)
Suy ra \(2A-A=2-2^{101}-1+2\Leftrightarrow A=3-2^{101}\)