2003 x 4 + 1998 + 2001 x 2002
------------------------------------------------
2002 + 2002 x 1002 + 2002 x 1003
tính nhanh
2003 x 14 + 1998 + 2001 x 2002
2002 + 2002 x + 503 + 504 x 2002
1 . Tính 2003 x 14 + 1998 + 2001 x 2002 phần 2002 + 2002 x 503 + 504 x 2002
cho mình sửa một chút 1998->1988
\(\frac{2003\times14+1998+2001\times2002}{2002+2002\times503+504\times2002}=\frac{\left(2002+1\right)\times14+1998+2001\times2002}{2002\left(1+503+504\right)}=\frac{2002\times14+14+1998+2001\times2002}{2002\times1008}\)=\(\frac{2002\times14+2002\times2001+2002}{2002\times1008}=\frac{2002\left(14+2001+1\right)}{2002\times1008}=\frac{2002\times2016}{2002\times1008}\)=\(\frac{2016}{1008}=\frac{2}{1}=2\)
phân số thì không ra mà chỉ ra số thập phân thôi nha bạn
Kết Quả là: 2,000004955 nha.
2003x14+1988+2001x2002 2002+2002x503+504x2002
Tính nhanh phân số :
2001 x 2002 + 2003 x 22 + 1980 / 2002 x 2003 - 2001 x 2002
sắp xếp :
\(\dfrac{1001}{2002};\dfrac{-1003}{2003};\dfrac{-1002}{2003};\dfrac{-1003}{-2002};\dfrac{1004}{-2003}\)
ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)
\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)
Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)
tính nhanh
2003 x 14 + 1988 + 2001 x 2002
2002 + 2002 x 503 + 504 x 2002
P=\(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}=\frac{2016}{1008}=2\)
(2001/2002 +2002/2003 + 2003/2004) x(1/3 - 1/4 - 1/12)
thực sự bạn có thể bấm máy tính đó đồ ngốc, ahihi
Với lại đây là toán hsg, vào phòng thi ngta cho mang máy tính à
S = 1 - 1/2 + 1/3 - 1/4 + ... + 1/2001 - 1/2002 P = 1/1002 + 1/1003 + ... + 1/2002 Hỏi S - P = ?
S=\(\left(1+\frac{1}{2}+......+\frac{1}{2002}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{2002}\right)\)
=\(\left(1+\frac{1}{2}+.........+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+.........+\frac{1}{1001}\right)\)
=\(\frac{1}{1002}+\frac{1}{1003}+...........+\frac{1}{2002}=P\)
\(\Rightarrow S-P=0\)
Giải phương trình sau :
\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)
\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005
x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4
<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+
(x-2002/3 -1)+(x-2001/4 -1) =0
<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-
x-2005/3- x-2005/4 =0
<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0
<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)
<=>x=2005
Vậy pt có nghiệm là x=2005
\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)
\(\Leftrightarrow\dfrac{x-4}{2001}-1+\dfrac{x-3}{2002}-1+\dfrac{x-2}{2003}-1=\dfrac{x-2003}{2}-1+\dfrac{x-2002}{3}-1+\dfrac{x-2001}{4}-1\)
\(\Leftrightarrow\dfrac{x-2005}{2001}+\dfrac{x-2005}{2002}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2}-\dfrac{x-2005}{3}-\dfrac{x-2005}{4}=0\)
\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\right)=0\)
\(\Leftrightarrow x-2005=0\)
\(\Leftrightarrow x=2005\)
Vậy nghiệm của PT là \(x=2005\)
Tính nhanh A:
A = 2003 x 14 + 1988 + 2001 x 2002
2002 + 2002 x 503 + 504 x 2002