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\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Rightarrow x+\frac{3}{4}=\frac{10}{9}:\frac{5}{7}=\frac{10}{9}\times\frac{7}{5}=\frac{14}{9}\)
\(\Rightarrow x=\frac{14}{9}-\frac{3}{4}=\frac{56-27}{36}=\frac{29}{36}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Leftrightarrow x+\frac{3}{4}=\frac{14}{9}\)
\(\Rightarrow x=\frac{29}{36}\)
P/s tham khảo nha
Giúp em bài toán này với :
Bài 3: Tìm x :
b) X x \(\frac{1}{2}\)+ \(\frac{3}{2}\)x X = \(\frac{4}{5}\)
\(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(\Rightarrow x.1=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}\)
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 \(\frac{1991.1999}{1995.1995}< 1.\)
b) so sánh qua phân số trung gian \(\frac{h}{h+2}\)
ta có \(\frac{h+1}{h+2}>\frac{h}{h+2}^{\left(1\right)}\)
ta lại có \(\frac{h}{h+2}>\frac{h}{h+3}^{\left(2\right)}\)
từ (1) và (2)
\(\Rightarrow\frac{h+1}{h+2}>\frac{h}{h+3}\)
a) so sánh qua phân số trung gian \(\frac{200}{408}\)
ta có \(\frac{203}{408}>\frac{200}{408}^{\left(1\right)}\)
ta lại có \(\frac{200}{408}>\frac{200}{449}^{\left(2\right)}\)
từ (1) và (2)
\(\Rightarrow\frac{203}{408}>\frac{200}{449}\)
a,=7/20
b,=1/2
c,7/20 hoặc 9/25...còn nhiều lắm
học tốt nhé !
1/3 = 10/10
2/5 = 12/30
1/3 < 11/10 < 2/5
Chị chỉ giúp như zậy đc thôi !
Mong sao giúp được em là tốt rồi
12/24= 1/2 = 1x50/2x50= 50/100
1818/1515= 18/15= 6/5= 6x20/5x20=120/ 100
...nhớ
chọn đúng mk nhé