a) \(a^3+a^2b-a^2c-abc=a^2\left(a+b\right)-ac\left(a+b\right)=a\left(a+b\right)\left(a-c\right)\)
b) mk chỉnh lại đề
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
c) \(4-x^2-2xy-y^2=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)
d) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
ồ cuk dễ nhỉ
Nếu các bn thích thì ...........
cứ cho NTN này nhé !
a) \(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\)
Vậy...
b) \(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\)\(3\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
Vậy...
c) \(x\left(x-1\right)+2x-2=0\)
\(\Leftrightarrow\)\(x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Vậy...
d) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\)\(x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow\)\(\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(2x+3=0\) (do x2 + 1 > 0 )
\(\Leftrightarrow\)\(x=-\frac{3}{2}\)
Vậy...