rút gọn biểu thức :
A=1+1/2+1/22+1/23+...+1/22012
so sánh :
A=2010+1/2010-1 và B=2010-1/2010-3
rút gọn biểu thức: P=\(\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
Rút gọn biểu thức:
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=-3^{2010}\)
So sánh hai biểu thức A và B biết rằng:
A= 2010/2011+2011/2012+2012/2010
B= 1/3 + 1/4 + 1/5 + ... + 1/17
A= 3,000...=3
B=1,0928...=1
Vậy A > B.
K mik nha. Mik làm đầu tiên mà.
Tính giá trị biểu thức
a, \(A=2010^{2010}.\left(7^{10}:7^8-3.16-2^{2010}:2^{2010}\right)\)
b, \(B=\left(\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}\right):\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right)+1:\left(30.1009-160\right)\)
a,
A = 20102010.[710:78-3.16-22010:22010]
= 20102010.[72-48-1]
= 20102010.0 = 0
b,
B = 1
\(A=2010^{2010}.\left[7^{10}:7^8-3.16-2^{2010}:2^{2010}\right]\)
\(A=2010^{2010}.\left[7^2-48-1\right]\)
\(A=2010^{2010}.0\)
\(Vay\)\(A=0\)
A= 20102010(72 - 48 - 1)
A=20102010(49-48-1)
A=20102010.0
A=0
Rút gọn
A=\(\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
??????????????????????????????????????????
Rú gọn biểu thức G = \(\frac{1^{2010}+2^{2010}+3^{2010}+....................+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+.....................+20^{2010}}\)
rút gọn biểu thức rồi tính giá trị
3(x-1)(x^2+x+1) +(x-1)^3-4x(x+1)(x-1) tại x =-2
(3xy-2)(9x^2y^2+6xy+4)-3xy(3xy+1)^2 tại x =-2010,y =-1/2010
rút gọn biểu thức :
G = \(\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)
\(C=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{1^{1010}.2^{2010}+2^{2010}.2^{2010}+2^{2010}.3^{2010}+...+2^{2010}.10^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{\left(1^{2010}+2^{2010}+3^{2010}+...+10^{2010}\right)+2^{2010}.2^{2010}.2^{2010}...2^{2010}}\)
\(=\dfrac{1}{2^{2010}+2^{2010}+2^{2010}+...+2^{2010}}\)
\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+....+20^{2010}}\\ =\dfrac{1^{2010}+2^{2010}+...+10^{2010}}{2^{2010}\left(1^{2010}+2^{2010}+...+10^{2010}\right)}\\ =\dfrac{1}{2^{2010}}\)
Theo bài ra, ta có:
\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+....+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+....+20^{2010}}\)
\(\Rightarrow G=\dfrac{1^{2010}+2^{2010}+3^{2010}+....+10^{2010}}{2^{2010}\left(1^{1010}+2^{2010}+3^{2010}+....+10^{2010}\right)}\)
\(\Rightarrow G=\dfrac{1}{2^{2010}}\)
Vậy \(G=\dfrac{1}{2^{2010}}\)
So sánh A=20102011+1/20102012+1 và B=20102010+1/20102011+1
thì mới nói nếu dấu chia trừ mũ là xong
ý mà không được vậy mũ ra âm 1 à
ồ được bằng 1/2010
so sánh giá trị của biểu thức sau A=1+(1+2)+(1+2+3)+.......+(1+2+3+...+2012) và B=1×2012+2×2011+3×2010+....+2012×1
Xét biểu thức A
A= 1+(1+2) +....... +(1+2+3+...+2012)
A = 1+1+2+1+2+3+...+1+2+3+...+2012
A có 2012 số 1
có 2011 số 2
...
có 1 số 2012
A = 1 x2012 +2x2011+...+2012x1
mà B = 1 x2012 +2x2011+...+2012x1
nên A=B