rút gọn : T= { 1/2 +1 } . { 1/3 +1 } . { 1/4 +1 } ... { 1/98 +1 } . { 1/99 +1 }
Rút gọn biểu thức:
T= (1/2+ 1).(1/3+ 1) .(1/4+ 1)....(1/98+ 1). (1/99+ 1)
T= (1/2+ 1).(1/3+ 1) .(1/4+ 1)....(1/98+ 1). (1/99+ 1)
T= 3/2+4/3+5/4+...+99/98+100/99
T= 100/2
T= 50
T=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)...\left(\frac{1}{99}+1\right)\)=\(\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\frac{100}{99}\)=2*100=200
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right)...\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)
\(=\frac{3.4.5...100}{2.3.4....99}\)
\(=\frac{100}{2}=50\)
Rút gọn (1/99+2/98+3/97+...+99/1):(1/2+1/3+1/4+...+1/100)
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^
Rút gọn biểu thức sau:T={1/2+1}.{1/3+1}.{1/4+1}....{1/98+1}.{1/99+1}.
\(T=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(T=\frac{1}{2}.100\)
\(T=50\)
Rút gọn phân số sau :
B= [1/2 + 1/3 + 1/4 + ... + 1/100] / [ 99/1 + 98/2 + 97/3 + ...........+ 1/99]
Rút gọn
A= 2^100+2^99+2^98.....+2+1
B=3^100+3^99+3^98....+3+1
C=4^100+4^99+....+4+1
D=2^100- 2^99+....+2^2 - 2 + 1
E=3^100 - 3^99 + 3^98....- 3 +1
Thu gọn
M= 2 + 2^2 + 2^3 ....+ 2^100
Cho A =2+2^2+2^3+....2^100. Tìm số tự nhiên x sao cho A + 1 = 2x
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
Rút gọn S=101+100+99+98+...+3+2+1 :101-100+99-98+...+3-2+1
Rút gọn P biết P = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
rút gọn :\(\frac{101+100+99+98+.,.+3+2+1}{101-100+99-98+...+3-2+1}\)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\left(101+1\right).100:2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
\(=\frac{5050}{1+1+...+1+1}\)(51 chữ số 1)
= \(\frac{5050}{51}\)
rút gọn B=1/2 +(1/2)^2+(1/2)^3+(1/2)^4+.....+(1/2)^98+(1/2)^99
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=\left[1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\right]-\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\right]\)
\(\Rightarrow B=1-\left(\dfrac{1}{2}\right)^{99}\)