\(\left(x-4\right)\left(x-6\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4< 0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4>0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>6\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< 6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow4< x< 6\)

Mà x là số nguyên nên x=5

Để \(\left(x-4\right)\left(x-6\right)< 0\) thì \(\left(x-4\right)\) và \(\left(x-6\right)\) trái dấu

Hay \(x-4>0;x-6< 0\) => \(x>4;x< 6\Rightarrow x=\left\{5\right\}\)

hoặc \(x-4< 0;x-6>0\) => \(x< 4;x>6\Rightarrow x\in\varnothing\)

Vậy: Giá trị x nguyên thỏa mãn điều kiện là 5.

Ta có : (x + 2) + (x + 4) + (x + 6) +......+ (x + 100) = 2700

<=> (x + x + x + ..... + x) + (2 + 4 + 6 + ..... + 100) = 2700

<=> 50x + 2550 = 2700

=> 50x = 2700 - 2550

=> 50x = 150

=> x = 150 : 50

=> x = 3

( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ........+ ( x + 100 )

Xét các số hạng thứ 2 trong mỗi ngoặc trên là dãy số 2; 4; 6; ....; 100 

Nhận xét :  4 - 2 = 2; 6 - 4 = 2

Quy luật : Hai số liền nhau hơn kém nhau 2 đơn vị 

Từ 2 đến 100 có :    ( 100 - 2 ) : 2 + 1 = 50 ( ngoặc ) hay 50 số x

Ta có biểu thức :

( x + x + x + ....... + x ) + ( 2 + 4 + 6+ .......+ 100 ) = 5750 

x * 50 + ( 100 + 1 ) * 50 : 2                                    = 5750

x * 50 + 2525                                                        = 5750

x * 50                                                                   = 5750 - 2525

x * 50                                                                   = 3225 

x                                                                          = 3225 : 50

x                                                                          = 64,5

tin mik nha ! nhớ k đúng cho mik nhé 

a) x=4

\(A=\dfrac{x}{\left(x+2022\right)^2}=\dfrac{x}{x^2+4044x+2022^2}=\dfrac{1}{x+4044+\dfrac{2022^2}{x}}=\dfrac{1}{\left(x+\dfrac{2022^2}{x}\right)+4044}\le\dfrac{1}{2.\sqrt{x}.\sqrt{\dfrac{2022^2}{x}}+4044}=\dfrac{1}{2..\sqrt{\dfrac{x.2022^2}{x}}+4044}=\dfrac{1}{4044+4044}=\dfrac{1}{8088}\)-\(A_{max}=\dfrac{1}{8088}\Leftrightarrow x=2022\)

A)    \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)

           \(x=\dfrac{22}{15}\)

b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)

C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

\(15:\left(x+2\right)=3\)

\(\Rightarrow x+2=15:3=5\)

\(\Rightarrow x=5-2=3\)

15:(x+2)=3

nên x+2=5

hay x=3

`15  : (x+2) = 3`

`<=>x + 2 = 5`

`<=> x =3`

\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)