Tính nhanh:
\(\frac{2003\cdot14+1988+2005\cdot2002}{2002+1001\cdot1006+5060\cdot2002}\)
Những câu hỏi liên quan
\(\frac{2003\cdot14+1988+2001\cdot2002}{2002+2002\cdot503+504\cdot2002}\)các bạn lưu ý * = dấu nhân
(2002+1)x14+1988+2001x2002
2002x(1+503+504)
2002x14+14+1988+2001x2002
2002x1008
2002x14+2002+2001x2002
2002x1008
2002x(14+2001+1)
2002x1008
2002x2016
2002x1008
2
1
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2003 x 14 + 1988 + 2001 x 2002
=28042 + 1988 + 4006002
=4036032
2002 + 2002 x 503 +504 x 2002
=2002 x 1 + 2002 x 503 + 504 x 2002
=2002 x (1 + 503 + 504)
=2002 x 908
=1817816
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\(\frac{2003\cdot14+1988+2001\cdot2002}{2002+2002\cdot503+504\cdot2002}\)
CÁC BẠN LÀM NHANH GIÚP MÌNH NHÉ . BẠN NÀO XONG ĐẦU TIÊN MÌNH TICK CHO . LƯU Ý : DẤU * LÀ DẤU GẠCH CHÂN NHÉ
CẢM ƠN CÁC BẠN NHIỀU LẮM
frac{1998cdot1996+1997+1995}{1997cdot1996-1995cdot1996} frac{1994cdot1993-1992cdot1993}{1992cdot1993+1994cdot7+1986}frac{399cdot45+55cdot399}{1995cdot1996-1991cdot1995} frac{2006cdotleft(0.4-3:7.5right)}{2005cdot2006}frac{1978cdot1979+1980cdot21+1958}{1980cdot1979-1978cdot1979} frac{2.43cdot12300-24.3cdot1230}{45cdot20.1+55cdot28.9+4.5+3.3-55cdot5.37}frac{1996cdot1997+1998cdot3+1994}{1997cdot1999-1997cdot199...
Đọc tiếp
\(\frac{1998\cdot1996+1997+1995}{1997\cdot1996-1995\cdot1996}\) \(\frac{1994\cdot1993-1992\cdot1993}{1992\cdot1993+1994\cdot7+1986}\)
\(\frac{399\cdot45+55\cdot399}{1995\cdot1996-1991\cdot1995}\) \(\frac{2006\cdot\left(0.4-3:7.5\right)}{2005\cdot2006}\)
\(\frac{1978\cdot1979+1980\cdot21+1958}{1980\cdot1979-1978\cdot1979}\) \(\frac{2.43\cdot12300-24.3\cdot1230}{45\cdot20.1+55\cdot28.9+4.5+3.3-55\cdot5.37}\)
\(\frac{1996\cdot1997+1998\cdot3+1994}{1997\cdot1999-1997\cdot1997}\) \(\frac{2003\cdot14+1988+2001\cdot2002}{2002+2002\cdot503+504\cdot2002}\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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\(\frac{1}{2003\cdot2002}\)-\(\frac{1}{2002\cdot2001}\)-......\(\frac{1}{3\cdot2}\)-\(\frac{1}{2\cdot1}\)
1/2003.2002 - 1/2002.2001 - ... - 1/3.2 - 1/2.1
= 1/2003.2002 - (1/2002.2001 + ... + 1/3.2 + 1/2.1)
= 1/2002.2003 - (1/1.2 + 1/2.3 + ... + 1/2001.2002)
= 1/2002.2003 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/2001 - 1/2002)
= 1/2002.2003 - (1 - 1/2002)
= 1/2002.2003 - 2001/2002
= 1/2002.2003 - 2001.2003/2002.2003
= 1-2001.2003/2002.2003
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tính nhanh: 2003*14+1988+2001*2002/2002+2002*503+504*2002
tính nhanh
2003 x 14 + 1988 + 2001 x 2002
2002 + 2002 x 503 + 504 x 2002
P=\(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}=\frac{2016}{1008}=2\)
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Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Chứng minh rằng:
1996×1991×1989-1993×1988×1995=2005×2010×2003-2007×2002×2009