Bài 1:
a.\(y.\left(x-z\right)+7\left(z-x\right)\)
\(=y\left(x-z\right)-7\left(x-z\right)\)
\(=\left(y-7\right)\left(x-z\right)\)
b,\(27x^2\left(y-1\right)-9x^3\left(1-y\right)\)
\(=27x^2\left(y-1\right)+9x^3\left(y-1\right)\)
\(=\left(27x^2+9x^3\right)\left(y-1\right)\)
Bài 2
a.\(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\left(5-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-3\end{matrix}\right.\)
b.\(4x\left(x-2004\right)-x+2004=0\)
\(4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\left(4x-1\right)\left(x-2004\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2004=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0,25\\x=2004\end{matrix}\right.\)
c.\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-x-1=0\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
bài 1
a) y(x-z)+7(z-x)= y(x-z)-7(x-z)= (x-z)(y-7)
b) 27x2.(y-1)-9x3.(1-y)= 27x2.(y-1)+9x3.(y-1)= (y-1)(27x2-9x3)
bài 2
a) 5(x+3)+2x(x+3)=0
=(x+3)(5+2x)=0
\(\Leftrightarrow\)x+3=0 hoặc 5+2x=0
=>x=-3 hoặc x=\(\dfrac{-5}{2}\)
b)=4x(x-2014)-(x-2014)=0
= (x-2014)(4x-1)=0
\(\Leftrightarrow\)x-2014=0 hoặc 4x-1=0
=>x=2014 hoặc x= \(\dfrac{1}{4}\)
câu c) thấy kì kì, k biết làm
