Đáp án:
\(\begin{array}{l} a,\ x=-2\ \text{hoặc}\ x=\dfrac13\\ b,\ x=0\ \text{hoặc}\ x=±3\\ c,\ \text{x = 2 hoặc x = 0}\\ d,\ \text{x = 3 hoặc x = 1}\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ (5x-4)^2-49x^2=0\\ \Leftrightarrow (5x-4)^2=49x^2\\ \Leftrightarrow (5x-4)^2=(±7x)^2\\ ⇔\left[ \begin{array}{l}5x-4=7x\\5x-4=-7x\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=-2\\x=\dfrac13\end{array} \right.\\ \text{Vậy x = (- 2) hoặc x = $\dfrac13$}\\ b,\ 4x^3-36x=0\\ ⇔4x(x^2-9)=0\\ ⇔\left[ \begin{array}{l}4x=0\\x^2-9=0\end{array} \right.\ ⇔\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\\ \text{Vậy x = 0 hoặc x = ±3}\\ c,\ 1-(x^2-2x+1)=0\\ ⇔(x-1)^2=1\\ ⇔\left[ \begin{array}{l}x-1=1\\x-1=-1\end{array} \right.\ ⇔\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\\ \text{Vậy x = 2 hoặc x = 0}\\ d,\ (3x-5)^2-(x+1)^2=0\\ ⇔(3x-5)^2=(x+1)^2\\ ⇔\left[ \begin{array}{l}3x-5=x+1\\3x-5=-x-1\end{array} \right.\ ⇔\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\\ \text{Vậy x = 3 hoặc x = 1}\end{array}\)
chúc bạn học tốt!
