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\(2A=\frac{1.2+2.3+3.4+...+98.99}{1.2+2.3+3.4+...+98.99}\)
\(2A=1\)
\(A=\frac{1}{2}\)
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
Lời giải:
Xét số hạng tổng quát:
\(\frac{2n+1}{[n(n+1)]^2}=\frac{1}{n(n+1)}.\frac{2n+1}{n(n+1)}=\frac{n+1-n}{n(n+1)}.\frac{n+(n+1)}{n(n+1)}\)
\(=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n}+\frac{1}{n+1}\right)=\frac{1}{n^2}-\frac{1}{(n+1)^2}\)
Do đó:
\(S=\frac{3}{(1.2)^2}+\frac{5}{(2.3)^2}+....+\frac{2n+1}{[n(n+1)]^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{(n+1)^2}\)
\(=1-\frac{1}{(n+1)^2}\)
a,
\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\sqrt{2}\cdot\dfrac{\sqrt{2^5}}{1-\sqrt{9}}\)
\(=2^2-\left(\sqrt{3}\right)^2+\dfrac{\sqrt{2}\cdot\sqrt{2^5}}{1-3}=4-3+\dfrac{\sqrt{2^6}}{-2}=1+\dfrac{8}{-2}=1+\left(-4\right)=-3\)
b,
\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}=\dfrac{49}{50}\cdot\dfrac{49}{50}=\dfrac{49^2}{50^2}=\dfrac{2401}{2500}\)
b) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}+.......-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{99}{100}-2x=\dfrac{1}{2}\)
\(\Leftrightarrow-2x=\dfrac{1}{2}-\dfrac{99}{100}\)
\(\Leftrightarrow-2x=\dfrac{-49}{100}\)
\(\Leftrightarrow x=\dfrac{-49}{100}:\left(-2\right)\)
\(\Leftrightarrow x=\dfrac{49}{200}\)
Vậy ...............................
1) Đặt \(A=1.2+2.3+3.4+....+98.99\)
Ta có:\(3A=3.\left(1.2+2.3+3.4+....+98.99\right)\)
\(3A=1.2.3+2.3.3+3.4.3+....+98.99.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}=323400\)
Ta có:\(\frac{A.y}{1}=184800\Rightarrow y=184800:323400=\frac{4}{7}\)
2)Đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185,8\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\)
Tổng quát:\(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right)a}-\frac{1}{a\left(a+1\right)}\)
Ta có:
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{37.38.39}\)
\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\Rightarrow B=\frac{370}{741}:2=\frac{185}{741}\)
Khi đó \(A=\frac{185}{741}.1428+185,8=...........\) (tự tính ra)
(*)số ko đẹp mấy
a) S = 1.2 + 2.3 + 3.4 + ... + 99.100
S có thể được viết lại thành:
S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)
= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)
Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:
S = n(n+1)(2n+1)/6
Với n = 99, ta có:
S = 99.100.199/6 = 331650
Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:
S = n(n+1)/2
Với n = 98, ta có:
S = 98.99/2 = 4851
Do đó, S = 331650 - 4851 = 326799
b) B = 4924.12517.28−530.749.45529.162.748
B có thể được viết lại thành:
B = (4924.12517.28) / (530.749.45529.162.748)
B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)
B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529
B = 108 / 45529
c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101
C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101
C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)
C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)
C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)
d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018
D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+.........+\left(1+2+3+......+98\right)}{1.2+2.3+3.4+.............+98.99}\) \(A=\dfrac{1+3+6+................+4851}{2+6+12+..........+9702}\)
\(A=\dfrac{1+3+6+..........+4851}{1.2+2.3+2.6+........+2.4851}\)
\(A=\dfrac{1}{2}\)
Vậy\(A=\dfrac{1}{2}\)