Question

kto pomoże

Answer 1

7

założenia:

x²-4≠0

x≠2 lub x≠(-2)

((-2)/(x²-4)) -(3/(2-x))+(1/(x+2))=1

((-2)/(x²-4))+(3*(x+2)/(x²-4)+((x-2)/(x²-4))=1

(-2+3x+6+x-2)/(x²-4)=1

(4x+2)/(x²-4)=1

(4x+2-x²+4)/(x²-4)=0

(-x²+4x+6)=0

x²-4x-6=0

 

Δ=16+24=40

√Δ=4√10

x₁=(4-2√10)/2=2-√10

x₂=(4+2√10)/2=2+√10

 

8.

założenie

x-2≠0

x≠2

(x+1)/(x-2)≤-2 //*(x-2)²

(x+1)(x-2)≤-2(x-2)²

(x+1)(x-2)+2(x-2)²≤0

(x-2)(x+1+2x-4)≤0

(x-2)(3x-3)≤0

x=2 lub x=1

 

x∈<1;2)

 

9.

założenia:

x≠0

x≠(-1)

 

(x+2)/x -(x-2)/(x+1)=0

((x+2)(x+1)-(x-2)x)/x(x+1)=0

(x²+3x+2-x²+2x)/(x(x+1))=0

(5x+2)/(x(x+1))=0//(x(x+1))

5x+2=0

5x=(-2)

x=(-2/5)