`25<5^x<3125`

`->5^2<5^x<5^5`

`->2<x<5`

`->x=3;4`

Để \(\dfrac{11}{2x+3}\) nhận giá trị nguyên thì \(2x+3\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta có bảng sau:

Vậy để \(\dfrac{11}{2x+3}\) nhận giá trị nguyên thì \(x\in\left\{-7;-2;-1;4\right\}\)

Để \(\dfrac{11}{2x+3}\)là số nguyên khi:

2x+3ϵƯ(11)= {-1;1;-11;11}

Ta có bảng sau:

2x+3 x -1 1 -11 11 -2 -1 -7 4

⇒x ϵ {-2;-1;-7;4} 

Số cam còn lại sau lần bán thứ hai là:

    [10+1] x 2 = 22 [quả]

Số cam còn lại sau lần bán thứ nhất là: 

    [22+1] x 2= 46 [quả]

Số cam lúc đầu là:

   [46+1] x 2 = 94 [quả]

Đáp số : 94 quả

\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)

\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)

\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)

Vậy x = 5/19

\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

Vậy x = 1/2 hoặc x = -6

\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy x = 7 hoặc x = -1

\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)

\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)

\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)

Em nên gõ công thức trực quan để đề bài rõ ràng nhé

\(11-\dfrac{7}{4}:\left(\dfrac{3}{8}-\dfrac{2}{3}\right)\)

\(=11-\dfrac{7}{4}:\left(\dfrac{9-16}{24}\right)=11-\dfrac{7}{4}:\left(-\dfrac{7}{24}\right)=11-\dfrac{7}{4}.\left(-\dfrac{24}{7}\right)\)

\(=11-\left(-6\right)=11+6=17\)

\(3^x+3^{x+2}+3^{x+4}=2457\)

\(\Leftrightarrow3^x+3^2.3^x+3^4.3^x=2457\)

\(\Leftrightarrow3^x.\left(1+3^2+3^4\right)=2457\)

\(\Leftrightarrow3^x.91=2457\)

\(\Leftrightarrow3^x=2457:91\)

\(\Leftrightarrow3^x=27=3^3\)

\(\Leftrightarrow x=3\)

\(3^x+3^{x+2}+3^{x+4}=2457\)

\(\Leftrightarrow3^x+3^x.9+3^x.81=2457\)

\(\Leftrightarrow3^x\left(1+9+81\right)=2457\)

\(\Leftrightarrow3^x.91=2457\)

\(\Leftrightarrow3^x=27\Leftrightarrow3^x=3^3\Leftrightarrow x=3\)

\(\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{13}{15\cdot4}\\ =\dfrac{5}{2}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{13}{60}\\ =\dfrac{193}{60}\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

 \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

......

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )

Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

         \(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)

         \(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)

  .........................

         \(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)

=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

=> D <  1 - \(\dfrac{1}{10}\)

=>D < \(\dfrac{9}{10}\)

=> D < \(\dfrac{10}{10}\)

 Vậy D < 1