Question

Tìm nghiệm của đa thức:

A(x)=\(\dfrac{1}{2}x-\dfrac{3}{4}\left(x-2\right)\)

B(x)=\(\dfrac{1}{4}x^3-25x\)

C(x)=\(x^3-2x^2+3x-6\)

D(x)=\(2x^2-x+10\)

Answer 1

a: Đặt A(x)=0

=>1/2x-3/4x+3/2=0

=>-1/2x=-3/2

hay x=3

b: Đặt B(x)=0

\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-25\right)=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}x-5\right)\left(\dfrac{1}{2}x+5\right)=0\)

hay \(x\in\left\{0;10;-10\right\}\)

c: Đặt C(x)=0

\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)

=>x-2=0

hay x=2

d: Đặt D(x)=0

\(\Rightarrow2x^2-x+10=0\)

\(\text{Δ}=\left(-1\right)^2-4\cdot2\cdot10=-79< 0\)

DO đó: PTVN