Câu hỏi của Phùng Đức Tài

Question

mọi người làm giúp e bài 1 nhé đc bài 2 thì càng tốt ạ. cảm ơn mọi người

GV Nguyễn Trần Thành Đạt · Accepted Answer

$Bài.1:\
a,3x-9y=3\left(x-3y\right)\
b,x^2-5x=x\left(x-5\right)\
c,\left(x-3\right)\left(x-5\right)-\left(2x+1\right)\left(3-x\right)=\left(x-3\right)\left(x-5\right)+\left(x-3\right)\left(2x+1\right)\
=\left(x-3\right)\left(x-5+2x+1\right)=\left(x-3\right)\left(3x-4\right)\
d,3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\
e,3\left(x+5\right)-x^2-5x=3\left(x+5\right)-x\left(x+5\right)\
=\left(x+5\right)\left(3-x\right)$Câu trả lời: $Bài.1:\
a,3x-9y=3\left(x-3y\right)\
b,x^2-5x=x\left(x-5\right)\
c,\left(x-3\right)\left(x-5\right)-\left(2x+1\right)\left(3-x\right)=\left(x-3\right)\left(x-5\right)+\left(x-3\right)\left(2x+1\right)\
=\left(x-3\right)\left(x-5+2x+1\right)=\left(x-3\right)\left(3x-4\right)\
d,3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\
e,3\left(x+5\right)-x^2-5x=3\left(x+5\right)-x\left(x+5\right)\
=\left(x+5\right)\left(3-x\right)$

GV Nguyễn Trần Thành Đạt · Answer

$Bài.2:\
a,x^3-9x=0\
\Leftrightarrow x.\left(x^2-9\right)=0\
\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x+3=0\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=-3\x=3\end{matrix}\right.\
b,5x\left(x+2\right)-3\left(x+2\right)=0\
\Leftrightarrow\left(5x-3\right)\left(x+2\right)=0\
\Leftrightarrow\left[{}\begin{matrix}5x-3=0\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\x=-2\end{matrix}\right.\
c,x^2-7x=0\
\Leftrightarrow x\left(x-7\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=7\end{matrix}\right.$Câu trả lời: $Bài.2:\
a,x^3-9x=0\
\Leftrightarrow x.\left(x^2-9\right)=0\
\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x+3=0\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=-3\x=3\end{matrix}\right.\
b,5x\left(x+2\right)-3\left(x+2\right)=0\
\Leftrightarrow\left(5x-3\right)\left(x+2\right)=0\
\Leftrightarrow\left[{}\begin{matrix}5x-3=0\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\x=-2\end{matrix}\right.\
c,x^2-7x=0\
\Leftrightarrow x\left(x-7\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=7\end{matrix}\right.$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP