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So sánh: \(\frac{23}{48}< \frac{47}{92}\)(Nhân chéo tử này với mẫu kia bên nào có kết quả lớn hơn thì bên đó lớn hơn bạn nhekk)
Ta có \(\frac{23}{48}< \frac{23}{46}=\frac{46}{92}< \frac{47}{92}\)
Vậy \(\frac{23}{48}< \frac{47}{92}\)
Quy đồng tử số : 6/7 = 6 x20/7 x 20 = 120/140 . Vì 140 lớn hơn 137 nen 120/140 < 120/137 hay 6/7 < 120/137 .Vay 6/7 < 120/37.
a) 6/7 = 120/ 140
Vì 120/140 < 120/137 nên 6/7 < 120/137
b)18/75 = 6/25 ; 28/112 = 1/4
Vì 6/25 : 1/4 = 24/25 nên 18/75 < 28/112
c)Ta có: 1 - 17/20 = 3/20 ; 1 - 22/25 = 3/25
Vì 3/20 > 3/25 nên 17/20 < 22/25
mik nha
a)\(\frac{16.17-5}{16.16+11}=\frac{16.17-16+11}{16.16+11}\)\(=\frac{16.\left(17-1\right)+11}{16.16+11}=\frac{16.16+11}{16.16+11}=1\)
b) \(\frac{45.16-17}{28+45.15}=\frac{45.\left(15+1\right)-17}{28+45.15}\)\(=\frac{45.15+45-17}{28+45.15}=\frac{45.15+28}{28+45.15}=1\)
c) \(\frac{7256.4375-725}{3650+4375.7255}=\frac{\left(7255+1\right).4375-725}{3650+4375.7255}\)\(=\frac{7255.4375+4375-725}{3650+4375.7255}\)\(=\frac{7255.4375+3650}{3650+4375.7255}=1\)
Câu C nhớ sửa 725 thành 7255 nha !
Bài giải
\(a,\text{ }\frac{16\cdot17-5}{16\cdot16+11}=\frac{16\cdot16+16-5}{16\cdot16+11}=\frac{16\cdot16+11}{16\cdot16+11}=1\)
\(b,\text{ }\frac{45\cdot16-17}{28+45\cdot15}=\frac{45\cdot15+45-17}{45\cdot15+28}=\frac{45\cdot15+28}{45\cdot15+28}=1\)
\(c,\text{ }\frac{7256\cdot4375-725}{3650+4375\cdot7255}=\frac{4375\cdot7255+4375-725}{4375\cdot7255+3650}=\frac{4375\cdot7255+3650}{4375\cdot7255+3650}=1\)
Bài làm
c ) Ta có :
\(\frac{2017}{2018}< 1\)
\(\frac{12}{11}>1\)
\(\Rightarrow\frac{2017}{2018}< \frac{12}{11}\)
trả lời
a, quy đồng rồi so sánh
b,quy đồng rồi so sánh
c,phân số nào có tử nhỏ hơn mẫu khi so sành với phân số có tử lớn hơn mẫu đều bé hơn
d,quy đồng rồi so sánh
chắc vậy chúc bn học tốt
Bài 1:
Ta có:
\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)
\(\Leftrightarrow N< M\)
Vậy \(M>N.\)
Bài 2:
Ta có:
\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)
\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)
\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
\(\Leftrightarrow A>B\)
Vậy \(A>B.\)
Bài 3:
\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)
\(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)
\(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)
Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)
\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm
\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)
Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)
Bài 4:
\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)
Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)
\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)
\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)
Vậy \(\frac{1991.1999}{1995.1995}< 1.\)
ta thay : 1- 2005/2006 = 1/2006 ; 1 - 2006/2007 = 1/2007
vi 1/2006 > 1/2007 nen 2005/2006 < 2006/2007
bai nay dung 100 % . bai la phuong phap phan bu , ban co the lam phuong phap phan so trung gian , quy dong tu so hoac mau so
\(1-\frac{2005}{2006}=\frac{1}{2006};1-\frac{2006}{2007}=\frac{1}{2007}\)
Vì \(\frac{1}{2006}>\frac{1}{2007}\Rightarrow\frac{2005}{2006}< \frac{2006}{2007}\)
\(\frac{210}{209}>1;\frac{3}{2}>1;\frac{28}{28}=1;\frac{7}{8}<1;\frac{5}{6}<1;\frac{2120}{2121}<1\)
ta co ta co
\(\frac{210}{209}-\frac{1}{209}=1\) \(\frac{7}{8}+\frac{1}{8}=1\)
\(\frac{3}{2}-\frac{1}{2}=1\) \(\frac{5}{6}+\frac{1}{6}=1\)
vi \(\frac{1}{209}<\frac{1}{2}\) nen \(\frac{210}{209}<\frac{3}{2}\) \(\frac{2120}{2121}+\frac{1}{2121}=1\)
vi \(\frac{1}{6}>\frac{1}{8}>\frac{1}{2121}nen\frac{5}{6}<\frac{7}{8}<\frac{2120}{2121}\)
--->\(\frac{5}{6}<\frac{7}{8}<\frac{2120}{2121}<\frac{28}{28}<\frac{210}{209}<\frac{3}{2}\)
\(\frac{18}{75}=\frac{6}{25}\)
\(\frac{28}{112}=\frac{1}{4}=\frac{6}{24}\)
Vì 25>24 nên \(\frac{6}{25}< \frac{6}{24}\Leftrightarrow\frac{18}{75}>\frac{28}{112}\)
\(\frac{18}{75}=\frac{6}{25}\)
\(\frac{28}{112}=\frac{1}{4}\)
MÀ : \(\frac{1}{4}=\frac{6}{24}>\frac{6}{25}\)
\(\Rightarrow\frac{18}{75}< \frac{28}{112}\)