thuc hien phep tinh
a, A = \(\frac{2004\cdot37+2004+2\cdot2004+2004\cdot59+2004}{324\cdot321-201\cdot324-324+101-18\cdot324}\)
b,M =\(\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
A = 2004×37+2004+2×2004+2004×59+2004 324×321−201×324−324×101−18×324
tính nhanh 2004 . 37 + 2004 + 2 * 2004 + 2004 * 59 + 2004 / 324 * 321 - 201 * 324 - 324 * 101 - 18 * 324
\(\frac{2004.37+2004+2.2004+2004.59+2004}{324.321-201.324-101.324-18.324}\)
\(=\frac{2004.\left(37+1+2+59+1\right)}{324.\left(321-201-101-18\right)}\)
\(=\frac{2004.100}{324.1}\)
\(=\frac{200400}{324}\)
\(=\frac{16700}{27}\)
cái con THANH NGÂN ngu thế
A= 2004 x 37 + 2004 + 2 x 2004 + 2004 x 59 + 2004
324 x 321 - 201 x 324 - 324 x 101 - 18 x 324
tính nhanh
cho A + b = 12 tính 13 * a + 5 * b + 13 * b + 5 * a
b,a = 2004 * 37 + 2004 + 2 * 2004 + 2004 * 59 + 2004/ 324 * 321 - 2001 * 324 - 324 * 101 - 18 * 324
Thuc hien phep tinh:
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
Tígiasgias trị biểu thức :
\(A=\frac{1\cdot2004+2\cdot2003+3\cdot2002+...+2004\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+2004\cdot2005}\)
Thuc hien phep tinh:
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005
Giải:
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
\(\Leftrightarrow\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{2005}}+\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B-\dfrac{1}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{2006}}}{\dfrac{2}{3}}\)
\(\Leftrightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{\dfrac{3^{2005}-1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{3^{2005}-1}{2.3^{2005}}\)
Vậy ...
thuc hien cac phep tinh
1-2-3+4+5-6-7+...+2001-2002-2003+2004
so sánh:
a) \(\frac{n}{n-3}\)và \(\frac{n+1}{n+2}\)
b) \(\frac{2003\cdot2004-1}{2003\cdot2004}\)và \(\frac{2004\cdot2005-1}{2004\cdot2005}\)
a) Ta có: \(\frac{n}{n-3}\)có tử số lớn hơn mẫu số. \(\Rightarrow\frac{n}{n-3}>1\)
Ta lại có: \(\frac{\left(n+1\right)}{n+2}< 1\)( vì \(\frac{\left(n+1\right)}{n+2}\) có tử bé hơn mẫu)
\(\Rightarrow\frac{n}{n-3}>\frac{\left(n+1\right)}{n+2}\)
b)
Mà: \(\frac{2003.2004-1}{2003.2004}=1\)( Loại hai số giống nhau ở cả tử và mẫu: 2003 , 2004)
Còn: \(\frac{2004.2005-1}{2004.2005}=1\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}=\frac{2004.2005-1}{2004.2005}\)
P/s: Mình không chắc câu b) Nhé
Ta thấy : n > n - 3
=> \(\frac{n}{n-1}>1\)
Có : n + 1 < n + 2
=> \(\frac{n+1}{n+2}< 1\)
=> \(\frac{n}{n-3}>\frac{n+1}{n+2}\)