Question

BT: Tìm nghiệm của đa thức 

A. A(x)=4x-1.                                      E. F(x)=5-x^2.

B. B(x)=2x^2-8.                                   F. B(x)=(1/2-x)×(2x+1).

C. C(x)=x^2+1.                                    G. C(x)=3x^2-x.

D. D(x)=(3x-2)×(2x-3).                       H. D(x)= 1/2x^2+x^3

GIÚP MK VS NHA       

Answer 1

a) Đặt A(x)=0

\(\Leftrightarrow4x-1=0\)

\(\Leftrightarrow4x=1\)

hay \(x=\dfrac{1}{4}\)

Answer 2

b) Đặt B(x)=0

\(\Leftrightarrow2x^2-8=0\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Answer 3

c) Đặt C(x)=0

\(\Leftrightarrow x^2+1=0\)

mà \(x^2+1\ge1>0\forall x\)

nên \(x\in\varnothing\)

Answer 4

d) Đặt D(x)=0

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{3};\dfrac{3}{2}\right\}\)

Answer 5

e) Đặt F(x)=0

\(\Leftrightarrow5-x^2=0\)

\(\Leftrightarrow x^2=5\)

hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)

Answer 6

f) Đặt B(x)=0

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Answer 7

g) Đặt C(x)=0

\(\Leftrightarrow3x^2-x=0\)

\(\Leftrightarrow x\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\dfrac{1}{3}\right\}\)

Answer 8

h) Đặt D(x)=0

\(\Leftrightarrow\dfrac{1}{2}x^2+x^3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\\dfrac{1}{2}+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-\dfrac{1}{2}\right\}\)