\(a,5x\left(x-1\right)-x+1\)
\(=5x\left(x-1\right)-\left(x-1\right)\)
\(=\left(5x-1\right)\left(x-1\right)\)
\(b,2\left(x+5\right)-x^2-5x\)
\(=2\left(x+5\right)-x\left(x+5\right)\)
\(=\left(2-x\right)\left(x+5\right)\)
Phần d,e mình tìm x nha<3
\(d,x^2-2x-3=0\)
Cách 1:
\(x^2-3x+x-3=0\)
\(x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Cách 2:
\(x^2-2x-3=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{4}\\x-1=-\sqrt{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)e, \(2x^2+5x^2.3=0\)
\(\Leftrightarrow17x^2=0\)
\(\Rightarrow x=0\)
a)
\(5x\left(x-1\right)-x+1\)
= \(5x\left(x-1\right)-\left(x-1\right)\)
= \(\left(x-1\right)\left(5x-1\right)\)
b)
\(2.\left(x+5\right)-x^2-5x\)
= \(2.\left(x+5\right)-x\left(x+5\right)\)
= \(\left(x+5\right)\left(2-x\right)\)
c)
\(x^2-2xy-4x^2+y^2\)
= \(-2x^2-x^2-2xy+y^2\)
= \(\left(-2x^2-2xy\right)-\left(x^2-y^2\right)\)
= \(-2x\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
= \(\left(x+y\right)\left(-2x-x+y\right)\)
= \(\left(x+y\right)\left(-3x+y\right)\)
d) Cách 1:
\(x^2-2x-3\)
= \(\left(x^2-2x+1\right)-4\)
= \(\left(x-1\right)^2-2^2\)
= \(\left(x-1-2\right)\left(x-1+2\right)\)
= \(\left(x-3\right)\left(x+1\right)\)
Cách 2:
\(x^2-2x-3\)
= \(\left(x^2-x\right)-\left(x-1\right)-4\)
= \(x\left(x-1\right)-\left(x-1\right)-2^2\)
= \(\left(x-1\right)\left(x-1\right)-2^2\)
= \(\left(x-1\right)^2-2^2\)
= \(\left(x-1-2\right)\left(x-1+2\right)\)
= \(\left(x-3\right)\left(x+1\right)\)
e) \(2x^2+5x^2.3\) = \(x^2\left(2+15\right)\) = \(17.x^2\)
