Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\frac{1991.1999}{1995.1995}< 1.\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{8}\right):\frac{47}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{8}\right)\times\frac{3}{47}=2008\)
\(2009-\frac{41}{9}\times\frac{3}{47}-x\times\frac{3}{47}+\frac{133}{8}\times\frac{3}{47}=2008\)
\(2009-\frac{41}{141}-x\times\frac{3}{47}+\frac{399}{376}=2008\)
\(2009+(\frac{399}{376}-\frac{41}{141})-x\times\frac{3}{47}=2008\)
\((2009+\frac{869}{1128})-x\times\frac{3}{47}=2008\)
\(x\times\frac{3}{47}=2009+\frac{869}{1128}-2008\)
\(x\times\frac{3}{47}=1\frac{869}{1128}\)
\(x\times\frac{3}{47}=\frac{1997}{1128}\)
\(x=\frac{1997}{1128}:\frac{3}{47}\)
\(x=\frac{1997}{72}\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=\)2008
\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2009-2008\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)=1.\frac{47}{3}=\frac{47}{3}\)
\(\frac{82}{18}+x-\frac{133}{18}=\frac{47}{3}\)
\(x=\frac{282}{18}-\frac{82}{18}+\frac{133}{18}\)
\(x=\frac{333}{18}=\frac{37}{2}\)
Đáp số \(x=\frac{37}{2}\)
xin lỗi bn dấu nhân nó bị trùng với x nên mk thay dấu nhân thành dấu "." theo cách lớp 6 nha.
Nếu có chỗ nào sai thì mk xin lỗi các bạn và mong các bạn góp ý
*****Chúc bạn học giỏi*****
a, \(y-\frac{8}{3}=\frac{7}{5}:\frac{7}{3}\frac{7}{5}.\frac{3}{7}=\frac{3}{5}\)
\(y=\frac{3}{5}+\frac{8}{3}=\frac{9}{15}+\frac{40}{15}=\frac{49}{15}\)
Bài làm:
Xét: \(101\times102-101\times101-50-51=101\times\left(102-101\right)-101=101\times1-101=0\)
\(\Rightarrow\left(1+2+4+8+...+512\right)\times\left(101\times102-101\times101-50-51\right)=0\)
\(\Rightarrow\frac{\left(1+2+4+8+...+512\right)\times\left(101\times102-101\times101-50-51\right)}{2+4+8+16+...+1024+2048}=0\)
Học tốt!!!!
\(=\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{18x20}\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{20}\right)\\ =2x\frac{9}{20}\\ =\frac{9}{10}\)
a) Ta có :
\(\frac{7}{12}< \frac{x}{24}< \frac{2}{3}\)
\(\Rightarrow\frac{14}{24}< \frac{x}{24}< \frac{16}{24}\)
\(\Rightarrow14< x< 16\)
\(\Rightarrow x=15\)
Vậy x = 15
b) \(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=195\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+4+7+...+28\right)=195\)
\(\Rightarrow10x+145=195\)
\(\Rightarrow10x=195-145\)
\(\Rightarrow10x=50\)
\(\Rightarrow x=50:10\)
\(\Rightarrow x=5\)
Vậy x = 5
c) \(\left(x+0,5\right)+\left(x+1,5\right)+\left(x+2,5\right)=33\)
\(\Rightarrow\left(x+x+x\right)+\left(0,5+1,5+2,5\right)=33\)
\(\Rightarrow3x+4,5=33\)
\(\Rightarrow3x=33-4,5\)
\(\Rightarrow3x=28,5\)
\(\Rightarrow x=28,5:3\)
\(\Rightarrow x=9,5\)
Vậy x = 9,5
_Chúc bạn học tốt_
a, \(\frac{7}{12}\)\(< \)\(\frac{x}{24}\)\(< \)\(\frac{2}{3}\)
\(\frac{14}{24}\)\(< \)\(\frac{x}{24}\)\(< \)\(\frac{16}{24}\)
Số lớn hơn 14 và nỏ hơn 16 là : 15
\(\Rightarrow\)Vậy \(x\)= 15