X x 56=765+21{31-12]
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x+1+\frac{1}{6}+1+\frac{1}{12}+..+1+\frac{1}{90}=10\)
=> 2x + 8 + \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=10\)
=> 2x + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=10-8\)
\(2x+1-\frac{1}{10}=2\)
=> 2x + \(\frac{9}{10}=2\)
=> 2x = 2 - 9/10
=>2x = 11/10
=> x = 11/10 : 2
x = 11/20
2x+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=10
2x+1+1/6+1+1/12+1+1/20+1+1/30+1+1/42+1+1/56+1+1/72+1+1/90=10
2x+(1+1+1+1+1+1+1+1)+(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)=10
2x+8+(1-1/2+1/2-1/3+...+1/9-1/10)=10
2x+1-1/10=10-8
2x+9/10=2
2x=2-9/10
2x=11/10
x=11/10/2
x=11/20
\(x+\frac{13}{12}+\frac{21}{20}+...+\frac{91}{90}\)
\(=x+1+\frac{1}{12}+1+\frac{1}{20}+...+1+\frac{1}{90}\)
\(=x+\left(1+1+1+1+1+1+1\right)+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=x+7+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(=x+7+\frac{7}{30}\)
\(=x+7\frac{7}{30}\)
a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
a)
\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)
\(=\frac{7\cdot11}{6\cdot12}\)
\(=\frac{77}{72}\)
b)
\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=6+\frac{1}{2}-\frac{1}{8}\)
\(=6+\frac{3}{8}\)
\(=\frac{51}{8}\)
Chia thành...a và b nhé.
Bg
a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)
= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)
= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)
= \(\frac{7.11}{6.12}\) (chịt tiêu trên dưới)
= \(\frac{77}{72}\)
b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)
Có 6 số hạng (đếm)
= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
= \(6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
= \(6+\frac{1}{2}-\frac{1}{8}\)
= \(\frac{13}{2}-\frac{1}{8}\)
= \(\frac{51}{8}\)
Hơi dài....
\(x\times56=765+21\left(31-12\right)\)
\(x\times56=756+21\times19\)
\(x\times56=1155\)
\(x=1155:56\)
\(x=\frac{165}{8}=20,625\)