1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)

2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

​\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)

\(-\dfrac{11}{12}< \dfrac{5}{x}< -\dfrac{11}{15}\)

\(\Rightarrow\dfrac{-55}{60}< \dfrac{5}{x}< \dfrac{-44}{60}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{50}{60}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{5}{-6}\)

Vậy x = -6

`(17x -25) : 8 + 65 =9^2`

`=> (17x -25) : 8 =81 -65`

`=> (17x -25) : 8 =16`

`=>17x -25=16 xx 8`

`=>17x -25=128`

`=>17x=128 + 25`

`=>17x=153`

`=>x=153:17`

`=>x=9`

`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:

⇒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

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

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

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