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Ta có : \(\frac{2009}{987654321}< \frac{2010}{987654321}\)
\(\frac{2010}{24681357}>\frac{2009}{24681357}\)
\(\Rightarrow A=B\)
\(A=\frac{2009}{987654321}+\frac{2010}{246813579}\)
\(=\frac{2009}{987654321}+\frac{2009}{246813579}+\frac{1}{246813579}\)
\(B=\frac{2009}{987654321}+\frac{1}{987654321}+\frac{2009}{246813579}\)
Có \(\frac{1}{246813579}>\frac{1}{987654321}\)
Vậy A > B
Quy đồng
A=(2006.246813579+2007.987654321)/(246...
B=(2007.246813579+2006.987654321)/(246...
MS bằng nhau nên ta so sánh tử:
A'=2006.246813579+2007.987654321=2006.... +987654321
=2006.(246813579+987654321)+987654321
B'=2007.246813579+2006.987654321 =2006.246813579+246813579+2006.987654321
=2006.(246813579+987654321)+246813579
987654321>246813579
Nên A'>B'
Hay A>B
x+1234567890=1234567890-987654321
<=> x + 1234567890 = 246913569
<=> x = 246913569 - 1234567890
<=> x = -987654321
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}\)
\(=\frac{3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{7}{2.6}\)
\(=\frac{7}{12}\)
\(=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6}\)
\(=\frac{1\cdot7}{2\cdot6}\)
\(=\frac{7}{12}\)
a) nếu k =1000 thì biểu thưc = 950
b) số lớn nhất có 4 chữ số là:9999
=> k= 9999+50=10049
a:1000-10*5=1000-50=950
b:9999+(10*5)=10049
đúng thì k tui nhá
A = m x 3 + n x 4 + p x 2 + m + 2 x p
= m x 4 + n x 4 + p x 4
= (m + n + p) x 4
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
1.21932631e+18
1219326311126352690