`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

\(2^{x-1}=16\)

=>\(2^{x-1}=2^4\)

=>x-1=4

=>x=4+1=5

\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\\Rightarrow x-1=4\\ \Rightarrow x=5\)

Vậy: \(x=5\)

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

3AN=2CN

=>\(AN=\dfrac{2}{3}CN\)

=>\(AN=\dfrac{2}{5}AC\)

=>\(CN=\dfrac{3}{5}AC\)

CM=2BM

=>\(BM=\dfrac{1}{3}BC;CM=\dfrac{2}{3}BC\)

Vì \(CN=\dfrac{3}{5}AC\)

nên \(S_{MNC}=\dfrac{3}{5}\cdot S_{AMC}\)

=>\(S_{AMC}=30:\dfrac{3}{5}=50\left(cm^2\right)\)

Vì \(CM=\dfrac{2}{3}BC\)

nên BC=1,5CM

=>\(S_{ABC}=1,5\cdot S_{AMC}=1,5\cdot50=75\left(cm^2\right)\)

Cần gấp

p là số nguyên tố lớn hơn 3 nên p=3k+1 hoặc p=3k+2

Nếu p=3k+1 thì \(2p+1=2\left(3k+1\right)+1=6k+3=3\left(2k+1\right)⋮3\)

=>Loại

Vậy: p=3k+2

\(4p+1=4\left(3k+2\right)+1=12k+8+1=12k+9=3\left(4k+3\right)⋮3\)

=>4p+1 là hợp số

\(D=\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{\left(23+1\right)\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{7}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\\ =\dfrac{23\cdot47+47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{3\left(3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}\right)}\\ =\dfrac{23\cdot47+24}{23\cdot47+24}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

\(12⋮n-2\)

=>\(n-2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(n\in\left\{3;1;4;0;5;-1;6;-2;8;-4;14;-10\right\}\)

Bổ sung điều kiện (n \(\ne\) 2)

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