Question

Cho các đa thức \(f\left(x\right)=4x^2+3x-2\);\(g\left(x\right)=x^2+2x+3\);\(h\left(x\right)=x\left(5x-2\right)+8.\)

a)Tính \(f\left(\dfrac{-1}{2}\right)\).

b)Tìm x để f(x)+g(x)-h(x)=0.

c)Chứng tỏ đa thức g(x) không có nghiệm.

Answer 1

a/ \(f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)-2\)

\(=4\cdot\dfrac{1}{4}-\dfrac{3}{2}-2=1-\dfrac{3}{2}-2=-\dfrac{5}{2}\)

b/

\(f\left(x\right)+g\left(x\right)-h\left(x\right)=4x^2+3x-2+x^2+2x+3-5x^2+2x-8\)

\(=\left(4x^2+x^2-5x^2\right)+\left(3x+2x+2x\right)+\left(-2+3-8\right)\)

\(=7x-7\)

Ta có: \(f\left(x\right)+g\left(x\right)-h\left(x\right)=7x-7=0\)

\(\Leftrightarrow7x=7\Rightarrow x=1\)

Vậy để...............

c/ \(g\left(x\right)=x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\)

Vì \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+2\ge2\)

hay \(\left(x+1\right)^2+2>0\)

\(\Rightarrow g\left(x\right)\) vô nghiệm (đpcm)