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d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
\(\frac{2929-101}{2.1919+404}\)
\(=\frac{101.\left(29-1\right)}{101\left(2.19+4\right)}\)
\(=\frac{28}{38+4}\)
\(=\frac{28}{42}\)
\(=\frac{2}{3}\)
Ta có :
\(\frac{2929-101}{2.1919+404}\)
\(=\frac{29.101-101}{2.19.101+4.101}=\frac{101\left(29-1\right)}{101\left(2.19+4\right)}=\frac{28}{38+4}=\frac{28}{42}=\frac{2}{3}\)
Ủng hộ mk nha !!! ^_^
\(\frac{101.29-101.1}{2.101.19+101.4}\)
\(=\frac{101.\left(29-1\right)}{101.\left(2.19+4\right)}\)
\(=\frac{28}{38+4}\)
\(=\frac{28}{42}=\frac{2}{3}\)
\(B=\frac{2929-101}{2.1919+404}\)
\(B=\frac{29.101-101}{38.101+4.101}\)
\(B=\frac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
\(B=\frac{2}{3}\)
Bài giải
a) Ta có : \(\frac{4545+101}{6969-303}=\frac{45.101+101}{69.101-101.3}=\frac{101.\left(45+1\right)}{101.\left(69-3\right)}=\frac{101.46}{101.66}=\frac{23}{33}\)
b) Ta có : \(\frac{2929-101}{2.1919+404}=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(19.2+4\right)}=\frac{28}{42}=\frac{2}{3}\)
a)\(\frac{4545+101}{6969-303}\)= \(\frac{\left(4545:45\right)+101}{\left(6969:69\right)-303}\)= \(\frac{101+101}{101-303}\)=\(\frac{202}{-202}\)=-1
b)\(\frac{2929-101}{2.1919+404}\)= \(\frac{2929-101}{3838+404}\)=\(\frac{\left(2929:29\right)-101}{\left(3838:38\right)+404}\)=\(\frac{101-101}{101+404}\)=\(\frac{0}{505}\)=0
học tốt
2929-101
___________
2.1919+404
2929-101*1
=_____________
2.1919+101*4
2929-1*1
=____________
2.1919+1*4
2929-1
=__________
2.1919+4
2928
=_____
6.1919
bó tay
2/3