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\(\frac{x+5}{100}+\frac{x+5}{99}=\frac{x+5}{98}+\frac{x+5}{97}\)
\(\Leftrightarrow\frac{x+5}{100}+\frac{x+5}{99}-\frac{x+5}{98}-\frac{x+5}{97}=0\)
\(\Leftrightarrow\left(x+5\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
\(\Leftrightarrow x+5=0\) (Vì: \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\) )
\(\Leftrightarrow x=-5\)
\(\frac{x+5}{100}+\frac{x+5}{99}=\frac{x+5}{98}+\frac{x+5}{97}\)
\(\Rightarrow\frac{x+5}{100}+\frac{x+5}{99}-\frac{x+5}{98}-\frac{x+5}{97}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
Mà \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\)
\(\Rightarrow x+5=0\)
\(\Rightarrow x=-5\)
Vậy \(x=-5\)
\(A=1+3+3^2+3^3+...+3^{99}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(\text{}\text{}\text{}1+3^2+3^3+...+3^{99}\right)\)
\(\Rightarrow2A=3^{100}-1\Rightarrow A=\frac{3^{100}-1}{2}\)
a, \(A=1+5+5^2+5^3+....+5^{2014}+5^{2015}\Rightarrow5A=5+5^2+5^3+5^4+.....+5^{2015}+5^{2016}\)
\(\Rightarrow5A-A=\left(5+5^2+5^3+....+5^{2015}+5^{2016}\right)-\left(1+5+5^2+.....+5^{2014}+5^{2015}\right)\)
\(\Rightarrow4A=5^{2016}-1\Rightarrow A=\frac{5^{2016}-1}{4}\)
b,\(B=2^{100}-2^{99}+2^{98}-.....-2^3+2^2-2\Rightarrow2B=2^{101}-2^{100}+2^{99}-.....-2^4+2^3-2^2\)
\(\Rightarrow B+2B=\left(2^{101}-2^{100}+2^{99}-.....-2^4+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-.....-2^3+2^2-2\right)\)
\(\Rightarrow3B=2^{101}-2\Rightarrow B=\frac{2^{101}-2}{3}\)
Đặt A = 12 + 32 + 52 + ... + 972 + 992
Đặt B = 22 + 42 + 62 + ... + 982
Khi đó A + B = 12 + 22 + 32 + ... + 982 + 992
= 1.1 + 2.2 + 3.3 + ... + 98.98 + 99.99
= 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ... + 98(99 - 1) + 99(100 - 1)
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - (1 + 2 + 3 + ... + 99)
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - 99.(99 + 1):2
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - 5050
Đặt C = 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3 + 99.100.3
3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97) + 99.100.(101 - 98)
3C = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100
3C = 99.100.101
C = 99.100.101 : 3 = 333 300
Khi đó A+ B = C - 5050 = 333 300 - 5050 = 328 250
Lại có B = 22 + 42 + 62 + ... + 982
= 22(12 + 22 + 32 + ... + 492)
= 4(12 + 22 + 32 + ... + 492)
Đặt D = 12 + 22 + 32 + ... + 492
= 1.1 + 2.2 + 3.3 + ... + 49.49
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ... + 49(50 - 1)
= 1.2. + 2.3 + 3.4 + ... + 49.50 - (1 + 2 + 3 + 4 + ... + 49)
= 1.2. + 2.3 + 3.4 + ... + 49.50 - 49.(49 + 1) : 2
= 1.2 + 2.3 + 3.4 + ... + 49.50 - 1225
Khi đó : 1.2 + 2.3 + 3.4 + ... + 49.50
= (1.2.3 + 2.3.3 + ... + 49.50.3) : 3
= [1.2.3 + 2.3.(4 - 1) + ... + 49.50(51 - 48)] : 3
= (1.2.3 + 2.3.4 - 1.2.3 + ... + 49.50.51 - 48.49.50) : 3
= 49.50.51 : 3
= 41650
Khi đó D = 41650 - 1225 = 40425
Khi đó B = 40425 x 4 = 161700
Lại có : A + B = 328250
=> A + 161700 = 328250
=> A = 166550
Vậy 12 + 32 + 52 + ... + 972 + 992 = 166550
a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
c) \(\left(x+2\right)-\left(x+3\right)>0\)
\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\ge\dfrac{-7}{3}\)
TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\le5\)
TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
a) 5A = 5 + 5^2 + 5^3 + 5^4 +...+ 5^51
=> 5A - A = 4A = 5^51 - 1
=> A = \(\frac{5^{51}-1}{4}\)
b) 3B = 3^100 - 3^99 -...- 3
=> 3B - B = 2B = 3^100 - 2.3^99 + 1
=> B = \(\frac{3^{100}-2\times3^{99}+1}{2}\)
a, 1+5+52+.....+550
=> 5(1+5+52+.....+550)=5+52+53.....+551
=>4(1+5+52+.....+550)=551-1
=>1+5+52+.....+550=(551-1):4
b,399-398-...-3-1
=399-(398+...+3+1)
=399-(399-1):2
\(C=5^{100}-5^{99}-5^{98}-5^{97}-...-5^2-5\)
\(C=5^{100}-\left(5^{99}+5^{98}+5^{97}+...+5^2+5\right)\)
\(C=5^{100}-\left(5+5^2+...+5^{97}+5^{98}+5^{99}\right)\)
\(\text{Đặt }B=5+5^2+...+5^{97}+5^{98}+5^{99}\)
\(5B=5^2+5^3+...+5^{98}+5^{99}+5^{100}\)
\(5B-B=5^{100}-5\)
\(4B=5^{100}-5\)
\(B=\frac{5^{100}-5}{4}\)
Thay \(B=\frac{5^{100}-5}{4}\) vào biểu thức ta có :
\(C=5^{100}-\frac{5^{100}-5}{4}\)
\(C=5^{100}-5^{99}-5^{98}-5^{97}-...-5^2-5\)
\(C=5^{100}-\left(5^{99}+5^{98}+5^{97}+...+5^2+5\right)\)
\(C=5^{100}-\left(5+5^2+...+5^{97}+5^{98}+5^{99}\right)\)
\(\text{Đặt }B=5+5^2+...+5^{97}+5^{98}+5^{99}\)
\(5B=5^2+5^3+...+5^{98}+5^{99}+5^{100}\)
\(5B-B=5^{100}-5\)
\(4B=5^{100}-5\)
Thay \(4B=5^{100}-5\) và \(3B=\frac{\left(5^{100}-5\right)}{3}\) vào biểu thức ta có :
\(C=5^{100}-\left(5^{100}-5\right)+\frac{\left(5^{100}-5\right)}{3}\)
\(C=5^{100}-5^{100}+5+\frac{5^{100}-5}{3}\)
\(c=5+\frac{5^{100}-5}{3}\)