Bn cứ áp dụng hằng đẳng thức đi
a)\(x^2-5x=6\)
\(\Rightarrow x^2-2.\dfrac{5}{2}.x+\dfrac{25}{4}=6+\dfrac{25}{4}\)
\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2=12,25\)
\(\Rightarrow x-\dfrac{5}{2}\in\left\{\pm3,5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=3,5\\x-\dfrac{5}{2}=-3,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
Vậy....
b)\(4x^2-20x+25=0\)
\(\Rightarrow4\left(x^2-5x\right)+25=0\)
\(\Rightarrow4\left(x^2-2.\dfrac{5}{2}.x+\dfrac{25}{4}\right)+25-25=0\)
\(\Rightarrow4.\left(x-\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{5}{2}=0\)
\(\Rightarrow x=\dfrac{5}{2}\)
c) Bn nhân đa thức vs đa thức r dùng hằng đẳng thức tính ra nha
c) \(\left(3x-1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-1-x+2\right)\left(3x-1+x-2\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy pt có 2 nghiệm là....
