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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
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
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{98}{2^{98}}+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\) (lấy 2A - A = A)
Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
\(B=2B-B=2-\frac{1}{2^{99}}\)
Do đó: \(A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)
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.
\(\frac{A}{2}=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{56}{305}\)
\(A=\frac{112}{305}\)
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=\(\frac{2^{101}-2}{3}\)