\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{45.47}\)

\(C=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)

\(C=\frac{5}{2}.\left(1-\frac{1}{47}\right)\)

\(C=\frac{5}{2}.\frac{46}{47}\)

\(C=\frac{115}{47}\)

= 205/101 nha

\(C=\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{45\cdot47}\)

\(C=\frac{5}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{45\cdot47}\right)\)

\(C=\frac{5}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)

\(C=\frac{5}{2}\cdot\left(1-\frac{1}{47}\right)\)

\(C=\frac{5}{2}\cdot\frac{46}{47}\)

\(C=\frac{115}{47}\)

100=25 .4 =5^2.4

5^2019:( 5^2013-5^2.4.5^2010)=5^2019: ( 5^2013-5^2012.4)=5^2019: (5^2012.(5-4))

=5^2019:5^2012=5^7(=78125)

Học tốt

5^x = 5²⁰¹⁹ : (5²⁰¹³ -100*5²⁰¹⁰)

<=> 5^x = 5²⁰¹⁹ : (5²⁰¹⁰ .(5³-100))

<=> 5^x = 5²⁰¹⁹ : (5²⁰¹⁰ . 25)

<=> 5^x = 5²⁰¹⁹ : 5²⁰¹²

<=> 5^x = 5⁷

<=> x = 7

Vậy x = 7

5^x=5²⁰¹⁹:(5²⁰¹³-25*4*5²⁰¹⁰)

5^x=5²⁰¹⁹:(5²⁰¹²*5-5²⁰¹²*4)

5^x=5²⁰¹⁹:5²⁰¹²

5^x=5⁷

Vậy x=7

làm theo ý nghĩ của bạn nhé BYE

Đề bài hơi sai thì phải

Mình cũng bó tay luôn. Rắc rối quá

Ta dùng phương pháp triệt tiêu sẽ được kết quả cuối cùng là : 
1 - \(\frac{1}{15}\) = \(\frac{14}{15}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(A=1-\frac{1}{15}\)

\(A=\frac{14}{15}\)

thank's các bạn nha

áp dụng tc \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{a+m}< 1\left(m\in N\right)\)

Ta có: \(A=\frac{5^{2010}+1}{5^{2011}+1}< \frac{5^{2010}+1+4}{5^{2011}+1+4}\)\(=\frac{5^{2010}+5}{5^{2011}+5}=\frac{5.\left(5^{2009}+1\right)}{5.\left(5^{2010}+1\right)}=\frac{5^{2009}+1}{5^{2010}+1}\)

\(\Rightarrow A< B\)

#)Giải : 

Đầu tiên ta so sánh : 

5²⁰¹⁰ và 5²⁰⁰⁹ 

Vì 2010 > 2009 => 5²⁰¹⁰ > 5²⁰⁰⁹    (1)

Tiếp theo :

1/5^{2011 + 1} và 1/5^{2010 + 1}

Vì 2011 + 1 = 2012 và 2010 + 1 = 2011 

Mà 2012 > 2011 => 1/5^{2011 + 1} > 1/5^{2010 + 1}   (2)

Từ (1) và (2) => 5²⁰¹⁰ + 1/5²⁰¹¹⁺¹ > 5²⁰⁰⁹+1/5²⁰¹⁰⁺¹ => A > B

Vậy : A > B

#)Nếu đúng thì bn bảo mk nha :D

      #~Will~be~Pens~#

Hoàng Nguyên Hiếu sai rồi 

\(5A=5^2+5^3+....+5^{2011}\)

\(5A-A=\left(5^2-5^2\right)+\left(5^3-5^3\right)+...+5^{2011}-5\)

4A = \(5^{2011}-5\)

A = \(\frac{5^{2011}-5}{4}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2009}+5^{2010}\right)\)

\(A=30.1+30.5^2+30.5^4+....+30.5^{2008}\)

\(A=30.\left(1+5^2+5^4+....+5^{2008}\right)\)

Vậy chia hết cho 30

a: \(2x+5⋮x+1\)

=>\(2x+2+3⋮x+1\)

=>\(3⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

b: \(5x+9⋮x+2\)

=>\(5x+10-1⋮x+2\)

=>\(-1⋮x+2\)

=>\(x+2\in\left\{1;-1\right\}\)

=>\(x\in\left\{-1;-3\right\}\)

c: \(2x+11⋮x+3\)

=>\(2x+6+5⋮x+3\)

=>\(5⋮x+3\)

=>\(x+3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-2;-4;2;-8\right\}\)

d: \(4x+9⋮2x+1\)

=>\(4x+2+7⋮2x+1\)

=>\(7⋮2x+1\)

=>\(2x+1\in\left\{1;-1;7;-7\right\}\)

=>\(2x\in\left\{0;-2;6;-8\right\}\)

=>\(x\in\left\{0;-1;3;-4\right\}\)

e: \(6x+7⋮3x+1\)

=>\(6x+2+5⋮3x+1\)

=>\(5⋮3x+1\)

=>\(3x+1\in\left\{1;-1;5;-5\right\}\)

=>\(3x\in\left\{0;-2;4;-6\right\}\)

=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

g: \(10x+13⋮5x+1\)

=>\(10x+2+11⋮5x+1\)

=>\(11⋮5x+1\)

=>\(5x+1\in\left\{1;-1;11;-11\right\}\)

=>\(5x\in\left\{0;-2;10;-12\right\}\)

=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)