b,x2-5x+4=0
<=>x2-4x-x+4=0
<=>x(x-4)-(x-4)=0
<=>(x-4)(x-1)=0
=>x=4 HOẶC x=1
c,<=>3x2-15x=60+15x+3x2
<=>-30x=60
<=>x=-2
Bài 3: Tìm x
b) \(x^2-5x+4=0\)
\(\Rightarrow x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+4=0\)
\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2-\dfrac{9}{4}=0\)
\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2-\left(\dfrac{3}{2}\right)^2=0\)
\(\Rightarrow\left(x-\dfrac{5}{2}+\dfrac{3}{2}\right)\left(x-\dfrac{5}{2}-\dfrac{3}{2}\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=4\)
c) \(3x\left(x-5\right)=60+15x+3x^2\)
\(\Rightarrow3x^2-15x=60+15x+3x^2\)
\(\Rightarrow3x^2-3x^2-15x-15x=60\)
\(\Rightarrow-30x=60\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
\(b) x^2 -5x+4=0 \)
\(<=> x^2 -4x-x+4 =0 \)
\(<=> (x^2-x)-(4x-4)=0\)
\(<=> x(x-1)-4(x-1)=0 \)
\(<=> (x-1)(x-4)=0\)
<=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy x = 1 hoặc x = 4
c) \(3x(x-5) = 60+15x+3x^2\)
\(<=> 3x^2 -15x = 60 +15x+3x^2\)
\(<=> 3x^2 - 3x^2-15x-15x=60 \)
\(<=> 30x = 60 <=> x = 60 : 30 = 2 \)
