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Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)
2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)
2A= 1-\(\dfrac{1}{99}\)
2A= \(\dfrac{98}{99}\)
A= \(\dfrac{98}{99}\) : 2
A=\(\dfrac{49}{99}\)
bài 1
\(2A=\left(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{99\cdot101}\right)\cdot2\)
\(=5\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\right)\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=5\left(1-\frac{1}{101}\right)\)
\(=5\cdot\frac{100}{101}\)
\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)
bài 2:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
2.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}.\left(\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow\)2x + 3 = 93
\(\Rightarrow\)2x = 93 - 3
\(\Rightarrow\)2x = 90
\(\Rightarrow\)x = 90 : 2 = 45
\(H=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{33.37}\)
= \(\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{37}\right)\)
= \(\frac{3}{4}\left(1-\frac{1}{37}\right)\)
= \(\frac{3}{4}.\frac{36}{37}=\frac{27}{37}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)
Bài 1: Bạn ơi số 2004 không thuộc dãy A
A có số số hạng là: (2005 - 5) : 4 + 1 = 501 (số hạng)
A = (2005 + 5) x 501 : 2 = 503505
Bài 2:
a) B = 4 . 1 + 4 . 5 + 4 . 52 + 4 . 53 + ... + 4 . 51000
=> B = 4 . ( 1 + 5 + 52 + 53 + .... + 51000)
b) 5 + 52 + 53 + .... + 51000 có tận cùng là 0 (Do các lũy thừa với cơ số là 5 thì có tận cùng là 5 [25] mà ở đây có số số hạng là chẵn)
=> 1 + 5 + 52 + 53 + .... + 51000 có tận cùng là số 1
=> 4 . ( 1 + 5 + 52 + 53 + .... + 51000) có tận cùng là 4.
Vậy B có tận cùng là 4.
Bài 3:
1. A = 1.3 + 3.5 + 5.7 + ......... + 97.99
=> A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + .... + 97.(97 + 2)
=> A = 12 + 1.2 + 32 + 3.2 + 52 + 5.2 + .... + 972 + 97.2
=> A = (12 + 32 + 52 + .... + 972) + (1.2 + 3.2 + 5.2 + .... + 97.2)
=> A = (12 + 32 + 52 + .... + 972) + 2(1 + 3 + 5 + .... + 97)
=> A = (12 + 32 + 52 + .... + 972) + 2 { (97 + 1) . [(97 - 1) : 2 + 1] : 2 }
=> A = (12 + 32 + 52 + .... + 972) + 24802
Đặt B = (12 + 32 + 52 + .... + 972)
=> B = 1.1 + 3.3 + 5.5 + .... + 97.97
=> B = 1.(0 + 1) + 3.(1 + 2) + 5.(4 + 1) + ..... + 97.(96 + 1)
=> B = 0 + 1.1 + 3 + 2.3 + 5 + 4.5 + .... + 97 + 96.97
=> B = (0 + 3 + 5 + .... + 97) + (1.1 + 2.3 + 4.5 + .... + 96.97)
=> B = 2400 + \(\frac{\left(97-1\right).97\left(97+1\right)}{6}\)
=> B = 2400 + 152096 = 154496
=> A = 154496 + 4802 = 159298
(Làm tương tự ở câu 2 nha)
Mình giải câu b) bài 1: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)\) (100 số hạng)
\(\Leftrightarrow100x+5050=5750\Leftrightarrow100x=5750-5050=700\)
Ta có: \(100x=700\Leftrightarrow x=7\)