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

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

\(\Rightarrow4< x< 6\)

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: \(x\times9,8-x:0.25=18,096\)

=>\(x\times9,8-x\times4=18,096\)

=>\(x\times\left(9,8-4\right)=18,096\)

=>x=18,096:5,8=3,12

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

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

vậy x=0 và x=2

3x(x-2)-x^2+2x=0

<=>3x(x-2)-x(x-2)=0

<=>(3x-x)(x-2)=0

<=>2x(x-2)=0

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2

giúp em  bài với ạ,em cảm ơn, em đang vội ạ

\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)

Bài 5:

a: 3x+2(5-x)=0

=>3x+10-2x=0

=>x+10=0

=>x=-10

b: 2x(x+3)+2(x+3)=0

=>(x+3)(2x+2)=0

=>(x+1)(x+3)=0

=>\(\left[\begin{array}{l}x+1=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-3\end{array}\right.\)

Bài 6:

\(A=2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)

\(=-6x^5+4x^4+2x^3-2x^2+2x^3-6x^2+2x\)

\(=-6x^5+4x^4+4x^3-8x^2+2x\)

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

mn trả lời đi mà,mình tick cho

5,2÷1,3