\(B=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

B = \(\dfrac{n+1}{n-3}\) ⇒ B = \(\dfrac{n-3+4}{n-3}\) ⇒ B = \(\dfrac{n-3}{n-3}\) + \(\dfrac{4}{n-3}\) ⇒ B = 1 + \(\dfrac{4}{n-3}\)

để B là số nguyên (n ϵ Z) thì 1 + \(\dfrac{4}{n-3}\) là số nguyên

 ⇒ \(\dfrac{4}{n-3}\) là số nguyên ⇒ n - 3 ϵ Ư(4) ⇒ n - 3 ϵ {-4;-2;-1;1;2;4}

⇒n ϵ { -1; 1; 2; 4; 5; 7}

\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) = \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)+ \(\dfrac{2}{4}\) ⇔ 1/x = 1/4; 2/y = 2/4 ⇔ x =y =4

a, \(M>0\Leftrightarrow\left\{{}\begin{matrix}x+5>0\\x+9>0\end{matrix}\right.\Leftrightarrow x>-5\)

.

sửa đề \(f\left(x\right)=4x^2-9\)

a, \(f\left(-2\right)=4.4-9=16-9=7\)

\(f\left(\dfrac{1}{4}\right)=\dfrac{4.1}{16}-9=\dfrac{1}{4}-9=-\dfrac{35}{4}\)

b, Ta có \(4x^2-9=-1\Leftrightarrow4x^2=8\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)

`f(x)=4^2-9`

Chỗ `4^2` có hơi sai đúng không bạn?

`(x+14).(x-4)`

`=>`\(\left[{}\begin{matrix}x+14=0\\x-4=0\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=0-14=-14\\x=0+4=4\end{matrix}\right.\)

Vậy `x={-14;4}`

`=>` Chọn `D`

.

a, \(A=x^2-2x+3\)Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Với x = 1/2 ta có \(A=\dfrac{1}{4}-1+3=\dfrac{1}{4}+2=\dfrac{9}{4}\)

Với x = -1/2 ta có \(A=\dfrac{1}{4}+1+3=\dfrac{1}{4}+4=\dfrac{17}{4}\)

b, Ta có \(\left|x\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Với x = 1/3 ta được \(B=\dfrac{1}{3}-3+\left|1-\dfrac{3.1}{3}\right|=\dfrac{1}{3}-3+8=\dfrac{1}{3}+5=\dfrac{16}{3}\)

Với x = -1/3 ta được 

\(B=-\dfrac{1}{3}-3+\left|1+\dfrac{3.1}{3}\right|=-\dfrac{1}{3}-3+2=-\dfrac{1}{3}-1=-\dfrac{4}{3}\)

a, A=x^2-2x+3A=x2−2x+3Ta có \left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.∣x∣=21​⇔⎣⎡​x=21​x=−21​​

Với x = 1/2 ta có A=\dfrac{1}{4}-1+3=\dfrac{1}{4}+2=\dfrac{9}{4}A=41​−1+3=41​+2=49​

Với x = -1/2 ta có A=\dfrac{1}{4}+1+3=\dfrac{1}{4}+4=\dfrac{17}{4}A=41​+1+3=41​+4=417​

b, Ta có \left|x\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.∣x∣=31​⇔⎣⎡​x=31​x=−31​​

Với x = 1/3 ta được B=\dfrac{1}{3}-3+\left|1-\dfrac{3.1}{3}\right|=\dfrac{1}{3}-3+8=\dfrac{1}{3}+5=\dfrac{16}{3}B=31​−3+∣∣​1−33.1​∣∣​=31​−3+8=31​+5=316​

Với x = -1/3 ta được 

B=-\dfrac{1}{3}-3+\left|1+\dfrac{3.1}{3}\right|=-\dfrac{1}{3}-3+2=-\dfrac{1}{3}-1=-\dfrac{4}{3}B=−31​−3+∣∣​1+33.1​∣∣​=−31​−3+2=−31​−1=−34​