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mk chỉnh lại đề
\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{99}{100}\)
\(=\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.\frac{5^2-1}{5^2}....\frac{10^2-1}{10^2}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{9.11}{10^2}\)
\(=\frac{2.3.4...9}{3.4.5...10}.\frac{4.5.6...11}{3.4.5...10}\)
\(=\frac{2}{10}.\frac{11}{3}=\frac{11}{15}\)
\(x+3\frac{1}{2}+x=24\frac{1}{4}\)
\(2x=24+\frac{1}{4}-3-\frac{1}{2}\)
\(2x=20+1+\frac{1}{4}-\frac{2}{4}\)
\(2x=20+\frac{3}{4}\)
\(2x=\frac{83}{4}\)
\(x=\frac{83}{4}:2\)
\(x=\frac{83}{8}\)
\(x+3\frac{1}{2}+x=24\frac{1}{4}\)
\(\Leftrightarrow x+\frac{7}{2}+x=\frac{97}{4}\)
\(\Leftrightarrow x+x=\frac{97}{4}-\frac{7}{2}\)
\(\Leftrightarrow2x=\frac{83}{4}\)
\(\Leftrightarrow x=\frac{83}{8}\)
a, thì dễ rồi bạn tự làm nhé
mk làm câu b thôi
b,\(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+......+\(\frac{1}{99x100}\)
= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{99}\)- \(\frac{1}{100}\)
= 1 - \(\frac{1}{100}\)
= \(\frac{99}{100}\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)
\(B=1\times\frac{1996}{1993}\times\frac{1993}{1995}=\frac{1996}{1995}\)
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
a) (58 + x) : 5 = 18
58 + x = 18 x 5
58 + x = 90
x = 90 - 58
x = 32
d) 320 : x - 10 = 5 x 48 : 24
320 : x - 10 = 10
320 : x = 10 + 10
320 : x = 20
x = 320 : 20
x = 16
c) (1/15 + 1/35 + 1/63) . x = 1
1/9 . x = 1
x = 1 : 1/9
x = 9
Dãy số 1, 2, 3,. .., 150 có 150 số.
Trong 150 số có
+ 9 số có 1 chữ số
+ 90 số có 2 chữ số
+ Các số có 3 chữ số là: 150 – 9 – 90 = 51 (chữ số)
Dãy này có số chữ số là:
1 x 9 + 2 x 90 + 3 x 51 = 342 (chữ số)
Đáp số: 342 chữ số
\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{24}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+...+\left(\dfrac{1}{22}+\dfrac{1}{23}\right)+\dfrac{1}{24}\)
\(=\dfrac{3+2}{2\times3}+\dfrac{5+4}{4\times5}+\dfrac{7+6}{6\times7}+...+\dfrac{23+22}{22\times23}+\dfrac{1}{24}\)
\(=\dfrac{5}{6}+\dfrac{9}{20}+\dfrac{13}{42}+...+\dfrac{45}{506}+\dfrac{1}{24}\)
\(=\left(\dfrac{5}{6}+\dfrac{9}{20}\right)+\left(\dfrac{13}{42}+\dfrac{17}{72}\right)+...+\left(\dfrac{37}{342}+\dfrac{41}{420}\right)+\left(\dfrac{45}{506}+\dfrac{1}{24}\right)\)
\(=\dfrac{77}{60}+\dfrac{275}{504}+\dfrac{3013}{8580}+\dfrac{7409}{28560}+\dfrac{4927}{23940}+\dfrac{793}{6072}\)
\(=\left(\dfrac{77}{60}+\dfrac{275}{504}\right)+\left(\dfrac{3013}{8580}+\dfrac{7409}{28560}\right)+\left(\dfrac{4927}{23940}+\dfrac{793}{6072}\right)\)
\(=\dfrac{4609}{2520}+\dfrac{2493675}{4084080}+\dfrac{4075097}{12113640}\)
\(=\dfrac{9792714646+3269207925+1801192874}{5354228880}\)
\(=\dfrac{14863115445}{5354228880}\)
Thêm bước rút gọn nữa anh Phong ơi:
\(\dfrac{14863115445}{5354228880}=\dfrac{990874363}{356948592}\)