Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) = 2(1-1/2+1/2-1/3+...+1/19-1/20)
= 2(1-1/20)
= 2.19/20
= 19/10
b) = 7(1/2-1/3+1/3-1/4+...+1/6-1/7)
= 7(1/2 - 1/7)
= 7.5/14
= 5/2
c) = 1/2-1/5+1/5-1/8+...+1/14-1/17
= 1/2 - 1/17
= 15/34
Chúc bạn học tốt nhé
a)2/1.2+2/2.3+....+2/19.20
=2(1/1.2+1/2.3+....+1/19.20)
=2(1-1/2+1/2-1/3+.....-1/20)
=2(1-1/20)
2(19/20)=38/20=19/10
b)7/2.3+7/3.4+7/4.5+7/5.6+7/6.7
7(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)
7(1/2-1/3+1/3-1/4+.....-1/7)
7(1/2-1/7)
7(7/14-2/14)=7.5/14=35/14=5/2
c)3/2.5+3/5.8+3/8.11+3/11.14+3/14.17
1/2-1/5+1/5-1/8+......+1/14-1/17
1/2-1/17=17/34-2/34=15/34
#)Giải :
a)\(2^6.5^6=\left(2.5\right)^6=10^6\)
b)\(8^2.5^2=\left(8.5\right)^2=40^2\)
c)\(4^3.5^3=\left(4.5\right)^3=20^3\)
d)\(5^2.6^2.3^2=\left(5.6.2\right)^2=60^2\)
e)\(\frac{625^5}{25^8}=\frac{\left(25^2\right)^5}{25^8}=\frac{25^{10}}{25^8}=25^2\)
g)\(\frac{3^9}{7}.\frac{7^9}{3}=\frac{\left(3.7\right)^9}{7.3}=\frac{21^9}{21}=21^8\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Vậy \(x\in\left\{\pm7\right\}\)
a) Ta thấy:
\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{2}\cdot\frac{3}{5}=\frac{y}{3}\cdot\frac{3}{5}\)\(\Rightarrow\frac{3x}{10}=\frac{y}{5}\)
Mà \(\frac{y}{5}=\frac{z}{6}\) nên ta có biểu thức: \(\frac{3x}{10}=\frac{y}{5}=\frac{z}{6}\) ( 1 )
Biểu thức ( 1 ) tương đương với:
\(\frac{3x}{10}=\frac{3y}{15}=\frac{3z}{18}=\frac{3x+3y+3z}{10+15+18}=\frac{3\left(x+y+z\right)}{43}=\frac{3\cdot43}{43}=3\)
Khi đó:
\(\frac{3x}{10}=3\) \(\Rightarrow x=\frac{3\cdot10}{3}=10\)
\(\frac{3y}{15}=3\)\(\Rightarrow\frac{y}{5}=3\) \(\Rightarrow y=3\cdot5=15\)
\(\frac{3z}{18}=3\)\(\Rightarrow\frac{z}{6}=3\) \(\Rightarrow z=3\cdot6=18\)
a, Nhân cả hai vế cho 5, ta được: X/10 = Y/15
Tương tự ta có: Y/15 = Z/18
Do đó: X/10 = Z/18 (=Y/15)
Theo đề bài, ta có: (X+Y+Z)/(10+15+18) = 43/43 = 1
X/10=1 => X=10
Y/15=1 => Y=15
Z/18=1 => Z=18
Mk làm bài này trên cơ sở bài làm của bạn:
\(A=\frac{7}{3.7}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{10}{9.10}\)
\(A=\frac{8-1}{3.4}-\frac{10-1}{4.5}+\frac{12-1}{5.6}-\frac{14-1}{6.7}+\frac{16-1}{7.8}-\frac{18-1}{8.9}+\frac{20-1}{9.10}\)
\(A=\frac{8}{3.4}-\frac{1}{3.4}+\frac{12}{5.6}-\frac{1}{5.6}+\frac{14}{6.7}-\frac{1}{6.7}+\frac{16}{7.8}-\frac{1}{7.8}+\frac{18}{8.9}-\frac{1}{8.9}+\frac{20}{9.10}-\frac{1}{9.10}\)
\(A=\left(\frac{8}{3.4}+\frac{12}{5.6}+\frac{14}{6.7}+\frac{16}{7.8}+\frac{18}{8.9}+\frac{20}{9.10}\right)-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\left(\frac{2.4}{3.4}+\frac{2.6}{5.6}+\frac{2.7}{6.7}+\frac{2.9}{8.9}+\frac{2.10}{9.10}\right)-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(\frac{2}{3}+\frac{2}{5}+\frac{2}{6}+\frac{2}{8}+\frac{2}{9}\right)-\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(A=\left[\left(\frac{2}{3}+\frac{2}{9}\right)+\left(\frac{2}{6}+\frac{2}{8}\right)+\frac{2}{5}\right]-\frac{7}{30}\)
\(A=\left(\frac{8}{9}+\frac{7}{12}+\frac{2}{5}\right)-\frac{7}{30}\)
\(A=\left(\frac{160}{180}+\frac{105}{180}+\frac{72}{180}\right)-\frac{42}{180}\)
\(A=\frac{337}{180}-\frac{42}{180}\)
\(A=\frac{295}{180}=\frac{59}{36}\)
a)\(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{2}-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{1}{3}+\frac{1}{2}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\left(\frac{2}{6}+\frac{3}{6}\right)-\frac{1}{7}\)
\(=\frac{1}{7}.\frac{5}{6}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{1}{7}\)
\(=\frac{5}{42}-\frac{6}{42}=-\frac{1}{42}\)
ta có
\(-\frac{3^7\cdot2^8}{6^7}=-\frac{3^7\cdot2^7\cdot2}{6^7}=-\frac{6^7\cdot2}{6^7}=-2\)2
\(B=-\frac{3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}=-\frac{3\cdot7^3\cdot7+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}=-\frac{7^3\cdot\left(3\cdot7+1\right)}{7^3\cdot\left(7^2\cdot6-2\right)}=-\frac{5}{73}\)