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A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1
P/s: làm từng phần một
1.
\(2A=2^2+2^3+...+2^{101}\)
\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(A=2^{101}-2\)
2.
\(\frac{A}{2}=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{56}{305}\)
\(A=\frac{112}{305}\)
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+..+0
B=0
C=2^100-(2^99+2^98+2^97+...+1)
đặt D=2^99+2^98+2^97+...+1
=>D=2^100-1
=>C=2^100-(2^100-1)=1
a) \(M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(M=(100^-99^2)+(98^2-97^2)+...+(2^2-1^2)\)
\(=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)\)
\(=100+99+98+97+...+2+1\)
\(=\frac{100(100+1)}{2}=5050\)
b) \(N=(20^2-19^2)+(18^2-17^2)+...+(2^2-1^2)\)
\(=(20-19)(20+19)+(18-17)(18+17)+...+(2-1)(2+1)\)
\(=20+19+18+17+...+2+1=\frac{20(20+1)}{2}=210\)
c) \(P=(-1)^n(-1)^{2n+1}(-1)^{n+1}\)
\(P=(-1)^{n+2n+1+n+1}=(-1)^{4n+2}=(-1)^{2(2n+1)}=1\)
a: \(=\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)
=0
b: \(=\left(-1\right)^{100}+\left(-1\right)^{99}+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+...+\left(1-1\right)\)
=0
c: \(=1^{100}-1^{99}+1^{98}-1^{97}+...+1^2-1\)
=0
f: \(=3\cdot\sqrt{9-5}+7=3\cdot2+7=13\)
a)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(3.2\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+3^8.2^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+3^8.2^{10}.5}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
b) đặt 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=2101-2
=>A=\(\frac{2^{101}-2}{3}\)
thank kiu
thank kiu
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