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a) Ta có : \(\frac{35}{91}\)= \(\frac{5}{13}=\frac{x}{13}\Rightarrow x=5\)
b, ta có
<=>6 (9+x)=5 (13-x)
<=>54+6x=65-5x
<=>6x+5x=65-54
<=>11x=11
<=>x=1
\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\times x=1\)
\(\left(\frac{63}{945}+\frac{27}{945}+\frac{15}{945}\right)\times x=1\)
\(\left(\frac{63+27+15}{945}\right)\times x=1\)
\(\frac{105}{945}\times x=1\)
\(\frac{1}{9}\times x=1\)
\(x=1:\frac{1}{9}\)
\(x=1\times\frac{9}{1}\)
\(x=9\)
Vậy x=9
\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(1)
Đặt \(A=\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\)
\(A=31\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow2A=31\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(2A=31\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(2A=31\left(2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(2A=31\left(2-\frac{1}{13}\right)\)
\(2A=31.\frac{25}{13}\)
\(2A=\frac{775}{13}\)
\(\Rightarrow A=\frac{775}{13}:2\)
\(A=\frac{775}{26}\)
Thay vào (1) ta có:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
\(a,\left(x+\frac{5}{3}\right)\times\frac{9}{13}=\frac{2}{3}\\ \Rightarrow x+\frac{5}{3}=\frac{26}{27}\\ \Rightarrow x=-\frac{19}{27}\)\(b,43770\div x=560-434\\ \Rightarrow43770\div x=126\\ \Rightarrow x=\frac{7295}{21}\)\(c,x\div3\frac{1}{3}=2\frac{2}{5}+\frac{7}{10}\\ x\div3\frac{1}{3}=\frac{31}{10}\\ x=\frac{31}{3}\)
Ta có \(\frac{x}{13}=\frac{35}{31}\)
<=>31x=13.35
<=>31x=455
<=>x=\(\frac{455}{31}\)
x.31 = 13.35
31x = 455
x = 455:31
Vậy x = 445/31
NHớ k cho mình nhé!
Gọi tổng dãy số hạng trên là A
A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)
Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số
A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\)
A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)
A] x/13=35/91
B] 9+x/13-x
C] x + x chia 5 nhân 7.5+ x chia 2 nhân 9 =315
Các bạn giải cả ba câu nha!!!
a, \(\frac{x}{13}=\frac{35}{91}\)
\(\frac{x}{13}=\frac{5}{13}\rightarrow x=5\)
b, Cậu xem lại đb xem có nhầm ko
Nếu đb là \(9+x=13-x\)
\(\rightarrow2x=4\rightarrow x=2\)
c, Cậu cx xem lại đb ik
\(\frac{x}{13}\)= \(\frac{35}{91}\)
=> X x 91 = 35 x 13
=>X x 91 = 455
X = 455 : 91
X = 5