3+3/7-3/11+3/1001-3/13/9/1001-9/13+9/7-9/11+9
1,Tính
[(24*47-23) / (24+47+23)] * [(3+3/7-3/11+3/1001-3/13) / (9/1001-9/13+9/7-9/11+9)]
Tính nhanh:
A=24.47-23/24+47-23 . (3+3/7-3/11+3/1001-3/13)/(9/1001-9/13+9/7-9/11+9)
tính nhanh:
\(\dfrac{24.47-23}{24+47.23}.\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)
\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
tính giá trị biểu thức:
A= 24 . 47 - 23 / 24 + 47 - 23 . (3 + 3 + 3/7 - 3 /11 +3 / 1001 - 3/13) / ( 9/ 1001 - 9 /13 +9/7 - 9/ 11 +9 )
\(\frac{24.47-23}{24+47.23}\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
chỗ giữa 2 phân số là dấu nhân hay sao mà chả thấy dấu j thế?
Tính nhanh :
\(A=\dfrac{24.47-23}{24+47-23}.\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)
A=(24.47-23)/(24+47-23) . [3(1+1/7-1/11-1/13+1/1001)]/[9(1+1/7-1/11-1/13+1/1001)]
=1105/48 . 3/9 =1105/144
Tính giá trị biểu thức:
P = 24+47×23/24×47-23 ×
3+3/7-3/1001-3/13/9
1001-9/13+9/7-9/11+9
A=\(\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
Hỏi A=?
Tinh
a} A=24x47-23\24+47- 23 x 3+3\7-3\11=3\1001-3\13 \ 9\1001-9\13+9\7-9\11+9
b} M=1+2 +2 mũ 2 +2mu3+....+2mu2012\2mu2014-2
#)Giải ;
b) Đặt \(N=1+2+2^2+2^3+...+2^{2012}\)
\(\Rightarrow2N=2+2^2+2^3+2^4+...+2^{2013}\)
\(\Rightarrow2N-N=N=\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(\Rightarrow N=2^{2013}-1\)
Thay N vào M, ta có :
\(M=\frac{2^{2013}-1}{2^{2014}-2}\)
Thêm Cho pen
\(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
Phải tính hết nhé
a, \(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}\)
\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3}{9}\)
\(A=\frac{1105}{48}\cdot\frac{1}{3}=\frac{1105}{144}\)
Vậy A = \(\frac{1105}{144}\)
b, Đặt A = 1 + 2 + 2^2 + 2^3 + ... + 2^2012
2A = 2(1 + 2 + 2^2 + 2^3 + ... + 2^2012)
2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2012
2A - A = (2 + 2^2 + 2^3 + 2^4 + ... + 2^2012) - (1 + 2 + 2^2 + 2^3 + ... + 2^2012)
A = 2^2012 - 1
=> M = \(\frac{2^{2012}-1}{2^{2014}-2}=\frac{2^{2012}-1}{2^{2014}-2}=\frac{2^{2012}-1}{2.2.\left(2^{2012}-1\right).}=0,25\)
Vậy M = 0,25