Rút gọn: M=\(2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
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Rút gọn biểu thức
b) B=2^100-2^99+2^98-2^97+...+2^2-2
c) C=3^100-3^99+3^98-3^97+...+3^2-3+1
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
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Rút gọn A= 2^100-2^99+2^98-2^97+...+2^2-2
Ta có: A = 2100 - 299 + 298 - 297 + ... + 22 - 2 (gồm 100 hạng tử)
A = (2100 - 299) + (298 - 297) + ... + (22 - 2) (gồm 50 cặp)
A = 299(2 - 1) + 297.(2 - 1) + ... + 2(2 - 1)
A = 299 + 297 + .... + 2
22A = 22(299 + 297 + ... + 2)
4A = 2101 + 299 + ... + 23
4A - A = (2101 + 299 + ... + 23) - (299 + 297 + ... + 2)
3A = 2101 - 2
A = \(\frac{2^{101}-2}{3}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{201}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(3A=2^{201}-2\)
\(A=\frac{2^{201}-2}{3}\)
Rút gọn : A = 2^100 - 2^99 + 2^98 - 2^97 + ... + 2^2 - 2
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}-2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow3A=2^{201}-2\)
\(\Rightarrow\)\(\frac{2^{201}-2}{3}\)
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mình lấy cả 2 nick tích cho nguyên linh đấy
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Xem thêm câu trả lời
rút gọn b=2^100-2^99-2^98-2^97-....+2^2-2
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 = 2^101 - 2^100 + 2^99 -2^98 + ...+2^3 - 2 ^ 2 + 2^100 - 2^99 + 2^98 - ...+2^2 - 2
3A = 2^101 - 2
A = (2^101 - 2) / 3
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rút gọn
A= 2^100 -2^99 +2^98 - 2^97 +........+2^2 -2
Ta có: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{101}-2^{100}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+...+2^2-2\right)\)
\(\Leftrightarrow3A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
1.Rút gọn:
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow A+2A=2^{101}-2\)
\(A\left(1+2\right)=2^{101}-2\)
\(A.3=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)
\(\Rightarrow B+3B=3^{101}-3\)
\(B\left(1+3\right)=3^{101}-3\)
\(4B=3^{101}-3\)
\(B=\frac{3^{101}-3}{4}\)
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Rút gọn :
\(A=2^{100}-2^{99}+2^{98}-2^{97}.............+2^2-2\)
\(B=3^{100}-3^{99}+3^{98}-3^{97}.........+3^2-3+1\)
a, \(A=...\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\frac{2^{101}-2}{3}\)
b, tương tự a \(B=\frac{3^{101}+1}{4}\)
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Rút gọn:
a/ A=2^100-2^99+2^98-2^97+............+2^2-2
b/ B=3^100-3^99+3^98-3^97+..............+3^2-3+1
Ai nhanh nhất là đúng nhất mk tick cho
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
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Rút gọn:
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
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