Question

Tìm nghiệm của đa thức:

a.\(x^2-7x+12\).

b.\(2x^2-5x+2\).

c.\(P\left(x\right)=\left(x-3\right).\left(x+4\right)\).

d.\(Q\left(x\right)=\left(\dfrac{1}{2}x-1\right).\left(\dfrac{1}{2}-\dfrac{2}{3}\right)\).

e.\(-4x+3\).

g.\(x^2+4x-3\).

h.\(x^2+4x+5\).

i.\(2x^2-2x+3\).

Giúp mình với @nguyen thi vang , @Nhã Doanh

Answer 1

\(d.Q=\left(\dfrac{1}{2}x-1\right).\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=0\)

\(\Rightarrow\dfrac{1}{2}x-1=0\Rightarrow x=2\)

e. \(-4x+3=0\Rightarrow-4x=-3\Rightarrow x=\dfrac{4}{3}\)

g. \(x^2+4x-3=0\Rightarrow x^2+2.2x+4-7=0\)

\(\Rightarrow\left(x+2\right)^2-7=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{7}\\-2-\sqrt{7}\end{matrix}\right.\)

h.

\(x^2+4x+5=0\)

Ta có:

\(x^2+4x+5=x^2+2.x.2+4+1=\left(x+2\right)^2+1>0\)

=> đa thức vô nghiệm

Answer 2

i)\(2x^2-2x+3=0\)

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

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

Answer 3

a.

\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b.

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

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

c.

\(P\left(x\right)=\left(x+3\right).\left(x+4\right)\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

p/s: đến đây mới chợt nhận ra có điều bất thường , thiếu vế phải-.-

Answer 4

\(c,P_{\left(x\right)}=\left(x-3\right).\left(x-4\right)\)

Ta có: \(\left(x-3\right).\left(x-4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

Vậy \(x\in\left\{-4;-3\right\}\)