a) \(\frac{12x-2}{4x+1}=\frac{12x+3-5}{4x+1}=3-\frac{5}{4x+1}\)
Để f(x) là số nguyên thì 5 chia hết cho (4x+1)
----------lập bảng-------
suy ra x = { 0;1}
b, *f(x)> 0 
=> \(\hept{\begin{cases}12x-2>0\\4x+1>0\end{cases}}\Rightarrow\hept{\begin{cases}x>\frac{1}{6}\\x>-\frac{1}{4}\end{cases}}\Rightarrow x>\frac{1}{6}\)hoặc \(\hept{\begin{cases}12x-2< 0\\4x+1< 0\end{cases}\Rightarrow x< -\frac{1}{4}}\)

Suy ra f(x)>0 khi \(\orbr{\begin{cases}x>\frac{1}{6}\\x< -\frac{1}{4}\end{cases}}\)

*f(x)<0
=> \(\hept{\begin{cases}12x-2>0\\4x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{1}{6}\\x< -\frac{1}{4}\end{cases}}}\)(loại)   
hoặc \(\hept{\begin{cases}12x-2< 0\\4x+1>0\end{cases}\Rightarrow-\frac{1}{4}< x< \frac{1}{6}}\)

Vậy f(x) < 0 khi -1/4 <x<1/6

thanks b

Ta có 

+ f(1/2) -f(1/2) = 1/2 -1 =0 ( vô lí)

f(x) không xácđịnh.

Ta cod \(\hept{\begin{cases}f\left(x\right)>1\\f\left(x\right)=\frac{x+3}{x-2}\end{cases}}\)

<=> \(\frac{x+3}{x-2}>1\)

<=> \(\frac{x+3}{x-2}-1>0\)

\(\Leftrightarrow\frac{x+3-x-2}{x-2}>0\)

\(\Leftrightarrow\frac{1}{x-2}>0\)

<=> x - 2 > 0

<=> x = 2

Vậy x = 2

@@ Học tốt

Takigawa Miraii

Trả lời:

Ta có:\(f\left(x\right)=\frac{x+3}{x-2}\) \(\left(Đk:x\ne2\right)\)

Để\(f\left(x\right)>1\)

\(\Leftrightarrow\frac{x+3}{x-2}>1\)

\(\Leftrightarrow\frac{x+3}{x-2}-1>0\)

\(\Leftrightarrow\frac{x+3-x+2}{x-2}>0\)

\(\Leftrightarrow\frac{5}{x-2}>0\)

\(\Leftrightarrow x-2>0\)

\(\Leftrightarrow x>2\)(Thỏa mãn Đk: \(x\ne2\))

Vậy\(x>2\)thì hàm số\(f\left(x\right)=\frac{x+3}{x-2}>1\)

Hok tốt!

Good girl

H ms bt sai 

Làm lại

Ta có \(\hept{\begin{cases}f\left(x\right)=\frac{x+3}{x-2}\\f\left(x\right)>1\end{cases}}\)  ( x khác 2 )

<=> \(\frac{x+3}{x-2}>1\)

<=> \(\frac{x+3}{x-2}-1>0\)

\(\Leftrightarrow\frac{x+3}{x-2}-\frac{x-1}{x-2}>0\)

\(\Leftrightarrow\frac{5}{x-2}>0\)

<=> x - 2 > 0 ( do 5 >0 ) 

<=> x > 2 
Vậy x > 2

Cs bác nào thấy sai thì cứ nói hẳn ra nhá đừng tk sai như v !!!

a)  x khác 1

b) f(7)=\(\frac{3}{2}\)

c)\(\frac{x+2}{x-1}\)=\(\frac{1}{4}\)<=> 4(x+2)=x-1<=>x=-3

d) f(x)=\(\frac{x+2}{x-1}\)=\(\frac{x-1+3}{x-1}\)= 1+\(\frac{3}{x-1}\)

f(x) có giá trị nguyên <=> x-1 thuộc Ư(3) <=> x-1 thuộc {+1;+3}

e) f(x)>1 <=> 1+\(\frac{3}{x-1}\)> 1 <=> \(\frac{3}{x-1}\)> 0 <=> x-1 >0 <=> x>1

a: \(F\left(3\right)=3\left(3-2\right)=3\cdot1=3\)

\(\left[F\left(\dfrac{2}{3}\right)\right]^2=\left[\dfrac{2}{3}\cdot\left(\dfrac{2}{3}-2\right)\right]^2\)

\(=\left[\dfrac{2}{3}\cdot\dfrac{-4}{3}\right]^2=\left(-\dfrac{8}{9}\right)^2=\dfrac{64}{81}\)

\(G\left(-\dfrac{1}{2}\right)=-\left(-\dfrac{1}{2}\right)+6=6+\dfrac{1}{2}=\dfrac{13}{2}\)

b: F(x)=0

=>x(x-2)=0

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

c: F(a)=G(a)

=>\(a\left(a-2\right)=-a+6\)

=>\(a^2-2a+a-6=0\)

=>\(a^2-a-6=0\)

=>(a-3)(a+2)=0

=>\(\left[{}\begin{matrix}a-3=0\\a+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)