Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

HM

undefined

ai chỉ em 2 bài này vs ạ

NM
10 tháng 10 2021 lúc 14:38

\(1,\\ a,\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\\ 2,\\ a,\Leftrightarrow\left(x+5\right)^2=0\Leftrightarrow x=-5\\ b,\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ c,\Leftrightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\\ d,\Leftrightarrow\left(x-3\right)^3=0\Leftrightarrow x=3\\ e,\Leftrightarrow3x\left(x^2-2x+3\right)=0\\ \Leftrightarrow3x\left(x^2-2x+1+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2+2=0\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=0\)

\(f,\Leftrightarrow3x\left(x^2-4x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Bình luận (0)
LL
10 tháng 10 2021 lúc 14:39

Bài 1:

a) \(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

b) \(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

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

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

d) \(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow\left(x+5\right)^2=0\Rightarrow x=-5\)

b) \(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)

c) \(\Rightarrow\left(x-9\right)^2=0\Rightarrow x=9\)

d) \(\Rightarrow\left(x-3\right)^3=0\Rightarrow x=3\)

e) \(\Rightarrow3x\left(x^2-6x+9\right)=0\)

\(\Rightarrow3x\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

f) \(\Rightarrow3x\left(x^2-4x+4\right)=0\)

\(\Rightarrow3x\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Bình luận (0)
TD
10 tháng 10 2021 lúc 14:56

bài 1 

a) \(\Rightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

b) \(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\Rightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\Rightarrow x\left(x-7\right)+2\left(x-7\right)=0\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-7=0\Rightarrow x=7\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)

d) \(\Rightarrow2x\left(2x+3\right)-\left(2x+3\right)=0\Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}2x+3=0\Rightarrow x=\dfrac{-3}{2}\\2x-1=0\Rightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

 

 

 

 

 

Bình luận (0)
TD
10 tháng 10 2021 lúc 15:10

bài 2

a) \(\Rightarrow\left(x+5\right)^2=0\Leftrightarrow x=-5\)

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

c) \(\Rightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\)

d) \(\Rightarrow\left(x-3\right)^3=0\Rightarrow x=3\)

e) \(\Rightarrow3x\left(x^2-2x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}3x=0\Rightarrow x=0\\x^2-2x+3=0\left(VN\right)\end{matrix}\right.\)

f) \(\Rightarrow\left\{{}\begin{matrix}x-2=0\Rightarrow x=2\\x=0\end{matrix}\right.\)

Bình luận (0)

Các câu hỏi tương tự
HN
Xem chi tiết
DK
Xem chi tiết
CV
Xem chi tiết
TH
Xem chi tiết
4A
Xem chi tiết