Bài 1:
a) \(\left(x+1\right)^2-\left(x-3\right)\left(x+1\right)=5\\ \Rightarrow x^2+2x+1-x^2+2x+3-5=0\\ \Rightarrow4x-1=0\\ \Rightarrow x=\dfrac{1}{4}\)
b) \(\left(3x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=0\\ \Rightarrow9x^2-6x+1-9x^2+4=0\\ \Rightarrow-6x+5=0\\ \Rightarrow x=\dfrac{5}{6}\)
Bài 2:
a) \(x^2+y^2+13=4x-6y\\ \Rightarrow x^2+y^2+13-4x+6y=0\\ \Rightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\\ \Rightarrow\left(x-2\right)^2+\left(y+3\right)^2=0\)
ta thấy \(\left(x-2\right)^2\ge0;\left(y+3\right)^2\ge0mà\left(x-2\right)^2+\left(y+3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
b) \(x^2+2y^2-2xy-4y+4=0\\ \Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)=0\\ \Rightarrow\left(x-y\right)^2+\left(y-2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-2=0\\y=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Bài 1:
a: Ta có: \(\left(x+1\right)^2-\left(x-3\right)\left(x+1\right)=5\)
\(\Leftrightarrow x^2+2x+1-x^2+2x+3-5=0\)
hay \(x=\dfrac{1}{4}\)
b: Ta có: \(\left(3x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+4=0\)
hay \(x=\dfrac{5}{6}\)
