Question

xét dấu các biểu thức sau

a)f(x)=\(\frac{-4}{3x+1}-\frac{3}{2-x}\)

b)f(x)=4x^2_1

c)f(x)=\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\)

Answer 1

a)\(F\left(x\right)>0\) khi x thuộc \(\left(\frac{-9}{8};\frac{-1}{3}\right)\cup\left(2;-\infty\right)\)

b) ta có công thức ax²+bx+c=0 thì có a(x-x1)(x-x2)

với x là nghiệm của phương trình trên

vây f(x)>0 khi x thuộc\(\left(-\infty;\frac{-1}{2}\right)\cup\left(\frac{1}{2};+\infty\right)\)

c)f(x)>0 khi x thuộc \(\left(-2;\frac{-1}{2}\right)\cup\left(1:+\infty\right)\)

Answer 2

a) f (x) = \(\frac{-4}{3x+1}-\frac{3}{2-x}\)

= \(\frac{-4\left(2-x\right)-3\left(3x+1\right)}{\left(3x+1\right)\left(2-x\right)}=\frac{-8+4-9x-3}{\left(3x+1\right)\left(2-x\right)}\) = \(\frac{-5x-11}{\left(3x+1\right)\left(2-x\right)}\)
BXD : x \(\frac{-11}{5}\) \(\frac{-1}{3}\) 2
f(x) - 0 + \(||\) - \(||\) +

Vậy f(x) < 0 <=> x ∈ ( -∞ ; \(\frac{-11}{5}\) ) U (\(\frac{-1}{3}\) ; 2)
f(x) > 0 <=> x ∈ ( \(\frac{-11}{5}\); \(\frac{-1}{3}\) ) U (2 ; +∞)

b) f(x) = 4x² -1
f(x) = (2x-1)(2x+1)
2x-1 =0 <=> x = \(\frac{1}{2}\)
2x +1 =0 <=> x= \(\frac{-1}{2}\)

BXD : x \(\frac{-1}{2}\) \(\frac{1}{2}\)
f(x) + 0 - 0 +

f(x) >0 khi x ∈ ( -∞ ; \(\frac{-1}{2}\)) U ( \(\frac{1}{2}\); +∞)
f(x) <0 khi x ∈ ( \(\frac{-1}{2}\); \(\frac{1}{2}\))

c) f(x) = \(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\)
2x +1 = 0 <=> x= \(\frac{-1}{2}\)
x-1 =0 <=> x = 1
x+2 =0 <=> x = -2

BXD : x -2 \(\frac{-1}{2}\) 1
f(x) + \(||\) - 0 + \(||\) -

Vậy f(x) >0 khi x ∈ ( -∞ ;-2) U ( \(\frac{-1}{2}\) ; 1)
f(x)<0 khi x ∈ ( -2 ; \(\frac{-1}{2}\)) U ( 1; +∞)