Question

cho hàm số f(x) = \(\frac{x+2}{x-1}\) a) Tính f(7) b) Tìm x để f(x) = \(\frac{1}{4}\) c) Tìm x để f(x) có giá trị nguyên d) Tìm x để f(x) > 1

Answer 1

a) Ta có:\(f\left(x\right)=\frac{x+2}{x-1}\)

f(7)=\(\frac{7+2}{7-1}=\frac{5}{6}\)

Vậy f(x)=5/6

b) Ta có: \(f\left(x\right)=\frac{x+2}{x-1}=\frac{1}{4}\)

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

=> 4(x+2)=1(x-1)

=> 4x+8=x-1

=> 4x-x=-1-8

=> 3x=-9

=>x=-3

Vậy để f(x)=1/4 thì x=-3

c) Để \(f\left(x\right)\in Z\Rightarrow\frac{x+2}{x-1}\in Z\)

=> x+2\(⋮x-1\)

=>(x+2)-(x-1)\(⋮x-1\)

=> x+2-x+1\(⋮x-1\)

=> 3\(⋮x-1\)

=> x-1\(\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

=> x\(\in\left\{2;0;4;-2\right\}\)

Vậy x \(\in\left\{2;0;4;-2\right\}\)

d)