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\(\frac{18}{75}=\frac{6}{25}\)
\(\frac{28}{112}=\frac{1}{4}=\frac{6}{24}\)
Vì 25>24 nên \(\frac{6}{25}< \frac{6}{24}\Leftrightarrow\frac{18}{75}>\frac{28}{112}\)
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) Ta có : \(\frac{12}{48}< \frac{12}{47}\); \(\frac{12}{48}< \frac{13}{48}\)
=> \(\frac{12}{48}< \frac{13}{47}\)
b) Ta có : \(\frac{7}{13}=1-\frac{6}{13}\)
\(\frac{17}{23}=1-\frac{6}{23}\)
Mà \(-\frac{6}{13}< -\frac{6}{23}\)=> \(\frac{7}{13}< \frac{17}{23}\)
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
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
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 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 \(\frac{1991.1999}{1995.1995}< 1.\)
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 \(\frac{23}{48}< \frac{47}{92}\)