\(x\cdot9,63-5,63\cdot x=15,6\)

=>\(x\left(9,63-5,63\right)=15,6\)

=>4x=15,6

=>x=15,6:4=3,9

1.2×x_4.68:1.3:1.2=0.6

`2,4 : (2-x) - 0,32 . 4,5 = 1,56`

`=> 2,4 : (2-x) - 1,44 = 1,56`

`=> 2,4 : (2-x) = 1,56 + 1,44`

`=> 2,4 : (2-x) = 3`

`=> 2 - x = 2,4 : 3`

`=> 2 - x = 0,8`

`=> x = 2 - 0.8`

`=> x = 1,2`

Vậy `x = 1,2`

\(2,4:\left(2-x\right)-0,32\cdot4,5=1,56\)

=>\(2,4:\left(2-x\right)=1,56+0,32\cdot4,5=3\)

=>2-x=2,4:3=0,8

=>x=2-0,8=1,2

a: \(\left(\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{73\cdot76}\right)\cdot x^2=2\dfrac{16}{19}\)

=>\(\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{73\cdot76}\right)\cdot x^2=2+\dfrac{16}{19}=\dfrac{54}{19}\)

=>\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\right)\cdot x^2=\dfrac{54}{19}\)

=>\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{76}\right)\cdot x^2=\dfrac{54}{19}\)

=>\(\dfrac{1}{3}\cdot\dfrac{18}{76}\cdot x^2=\dfrac{54}{19}\)

=>\(\dfrac{6}{76}\cdot x^2=\dfrac{54}{19}\)

=>\(x^2=\dfrac{54}{19}:\dfrac{6}{76}=\dfrac{54}{19}\cdot\dfrac{76}{6}=9\cdot4=36\)

=>\(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

b: \(2^x+2^{x+2}=\dfrac{200}{19}\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)

=>\(2^x+2^x\cdot4=\dfrac{200}{19}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

=>\(5\cdot2^x=\dfrac{200}{19}\left(1-\dfrac{1}{20}\right)=\dfrac{200}{19}\cdot\dfrac{19}{20}=10\)

=>\(2^x=2\)

=>x=1

Bạn xem lại đề, quy luật của các số hạng trong tổng có vẻ chưa rõ ràng lắm.

Gọi hai thừa số lần lượt là a;b.

a.b=276

(a+19).b=713

a.b+b.19=713

b.19=713-276

b.19=437

b=437:19

b=23

a=276:23

a=12

Vậy hai số đó là 12 và 23

 Gọi số phải tìm là: A

   A:72=C dư 49 => A=Cx72+49                                                                                                                                                                                                                                                          A:75=C dư 28 => A=Cx75+28       

   Vậy ta có Cx72+49=Cx75+28    

   75xC-72xC=49-28

       3xC        = 21

        C           = 21:3=7

          =>       =72x7+49 =553

Vậy số cần tìm là : 553

      A = 3 + 3² + 3³ + ... + 3²⁰¹⁵

     3A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶

3A - A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶

    2A    = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶ - (3 + 3² + 3³ + ... + 3²⁰¹⁵)

    2A =  3² + 3³ + 3⁴ + ... + 3²⁰¹⁶ - 3 - 3² - 3³ - ... - 3²⁰¹⁵

  2 A =  3 ²⁰¹⁶ - 3 + (3² - 3²) + (3³ - 3³) + ... + (3²⁰¹⁵ - 3²⁰¹⁵)

    2A =  3²⁰¹⁶ - 3 + 0 + 0 + ... + 0 + 0

    2A = 3²⁰¹⁶ - 3

2A + 3 = 3²⁰¹⁶ - 3 + 3 

 2A + 3 = 3²⁰¹⁶ - (3 - 3) = 3^4\(x\)

3²⁰¹⁶ = 3^4\(x\)

2016 = 4\(x\)

  \(x\) = 2016 : 4

  \(x=\) 504

Vậy \(x=504\)

`5(x+12)+22=92`

`5(x+12)=92-22`

`5(x+12)=70`

`x+12=70/5`

`x+12=14`

`x=14-12`

`x=2`

Vậy: ... 

\(5\cdot\left(x+12\right)+22=92\\ \Rightarrow5\cdot\left(x+12\right)=92-22\\ \Rightarrow5\cdot\left(x+12\right)=70\\ \Rightarrow x+12=70:5\\ \Rightarrow x+12=14\\ \Rightarrow x=14-12\\ \Rightarrow x=2\)

c; \(\dfrac{\left(0,3\right)^6.\left(0,04\right)^3}{\left(0,09\right)^4.\left(0,2\right)^4}\)

= \(\dfrac{0,3^6.0,2^6}{0,3^8,0,2^4}\)

=  \(\dfrac{0,04}{0,09}\)

= \(\dfrac{4}{9}\)

d; \(\dfrac{2^3+2^4+2^5+2^6}{15^2}\)

=  \(\dfrac{8+16+32+64}{225}\)

= \(\dfrac{\left(8+32\right)+\left(16+64\right)}{225}\)

= \(\dfrac{40+80}{225}\)

= \(\dfrac{120}{225}\) 

= \(\dfrac{8}{15}\)

\(1\cdot4+2\cdot5+...+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+...+100\left(100+3\right)\)

\(=3\left(1+2+...+100\right)+\left(1^2+2^2+...+100^2\right)\)

\(=\dfrac{3\cdot100\cdot101}{2}+\dfrac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)

\(=3\cdot50\cdot101+338350=353500\)