Question

Bài 16: Tìm các giá trị của biến số x để phân thức sau bằng không:

       a) \(\frac{2x+3}{4x-5}\)

b)\(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}\)

c)\(\frac{x^2-1}{x^2-2x+1}\)

d)\(\frac{x^2-4}{x^2+3x-10}\)

e)\(\frac{x^3-16x}{x^3-3x^2-4x}\)

g)\(\frac{x^3+x^2-x-1}{x^3+2x-3}\)

Answer 1

a) Để \(\frac{2x+3}{4x-5}=0\)

=> 2x + 3 = 0

x = -3/2

b) Để \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-3\right).\left(x-1\right)}=\frac{x+2}{x-3}=0\)

=> x + 2 = 0=> x = -2

c) để \(\frac{x^2-1}{x^2-2x+1}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-1\right)^2}=\frac{x+2}{x-1}=0\)

=> x + 2 = 0 => x = -2 

d) để \(\frac{x^2-4}{x^2+3x-10}=\frac{\left(x+2\right).\left(x-2\right)}{\left(x-2\right).\left(x+5\right)}=\frac{x+2}{x+5}=0\)

=> ...

e) để \(\frac{x^3-16x}{x^3-3x^2-4x}=\frac{x.\left(x-4\right).\left(x+4\right)}{x.\left(x-4\right).\left(x+1\right)}=\frac{x+4}{x+1}=0\)

=> ....