\(5\dfrac{6}{4453}.\dfrac{1}{1997}-\dfrac{2}{1997}.2\dfrac{3}{4453}\)
\(F=5\dfrac{6}{4453}.\dfrac{1}{1997}-\dfrac{2}{1997}.2\dfrac{3}{4453}\)
rút gọn bằng cách thay số bằng chữ
Đặt a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
Tính :
1 . F = 5\(\dfrac{6}{4453}\). \(\dfrac{1}{1997}\)- \(\dfrac{2}{1997}.2\dfrac{3}{4453}\)
2 . G = 4\(\dfrac{7}{5741}\).\(\dfrac{1}{3759}\) - \(\dfrac{4}{3759}\). 1\(\dfrac{2}{5741}\)+ \(\dfrac{1}{3759}\)+ \(\dfrac{1}{3759.5741}\)
\(F=5\frac{6}{4453}.\frac{1}{1997}-\frac{2}{1997}.2\frac{3}{4453}\)
\(5\frac{6}{4453}\)x \(\frac{1}{1997}\)- \(\frac{2}{1997}\)x \(2\frac{3}{4453}\)
Đặt \(a=\frac{1}{4453};b=\frac{1}{1997}\)ta có :
\(5\frac{6}{4453}\cdot\frac{1}{1997}-\frac{2}{1997}\cdot2\frac{3}{4453}\)
\(=\left(5+6a\right)\cdot b-2b\left(2+3a\right)\)
\(=5b+6ab-4b-6ab\)
\(=b=\frac{1}{1997}\)
Tính nhanh
F = \(5\frac{6}{4453}x\frac{1}{1997}-\frac{2}{1997}x2\frac{3}{4453}\)
M = \(\frac{1}{4587}x7\frac{1}{3897}-3\frac{4586}{4587}x\frac{2}{3897}-\frac{7}{4587}-\frac{3}{4587x3897}\)
x là dấu nhân
Tính thuận tiện:
a) \(1987+1992+1997+2002+2007\)
b) \(4,25+4,32+4,39+4,46+4,53\)
c) \(\dfrac{3}{5}+\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{5}{8}+\dfrac{1}{2}\)
a: =5991+3994=9985
b: =12,96+8,99=21,95
c:=3/5+2/5+3/4+5/8+1/2
=1+6/8+5/8+4/8
=1+15/8
=23/8
\(\dfrac{x+1}{1998}\)+ \(\dfrac{x+2}{1997}\)=\(\dfrac{x+3}{1996}\)+\(\dfrac{x+4}{1995}\)
=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)
=>x+1999=0
=>x=-1999
Sử dụng tính chất bắc cầu để so sánh các phân số sau:
a) \(\dfrac{1997}{1996}và\dfrac{1996}{1997}\)
b) \(\dfrac{3}{5}và\dfrac{15}{13}\)
\(a,\dfrac{1997}{1996}>1>\dfrac{1996}{1997}\\ b,\dfrac{3}{5}< 1< \dfrac{15}{13}\)
\(\dfrac{x}{y}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1329}-\dfrac{1}{1330}+\dfrac{1}{1331}\)
CM x⋮ 1997