a) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4=2^2\)
\(\Leftrightarrow x-3=2\)
\(\Leftrightarrow x=5\)
Vậy : x = 5
b) (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
=>4x2-4x+1+x2+6x+9+245-5x2=0
=>(4x2+x2-5x2)+(6x-4x)+(1+9+245)=0
=>2x+255=0
=>2x=-255 <=>x=-255/2
a/\(\left(x-3\right)^2-4=0\)
⇔ \(\left(x-3\right)^2=2^2\)
⇔ \(\left(x-3\right)=2\)
⇔ \(x=5\)
b/\(\left(2x-2\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
⇔\(4x^2-4x+1+x^2+6x+9+245-5x^2=0\)
⇔\(\left(4x^2+x^2-5x^2\right)+\left(6x-4x\right)+\left(1+9+245\right)=0\)
⇔ \(2x+255=0\)
⇔\(2x=-255\)
⇔\(x=\frac{-255}{2}\)
a, ( x -3)\(^2\) - 4 =0
\(\Leftrightarrow\) ( x -3)\(^2\) = 4
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-3=\sqrt{4}\\x-3=-\sqrt{4}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy x = \(\left\{5;1\right\}\)
b, ( 2x -1 )\(^2\)+ ( x+3 )\(^2\) - 5( x+7)(x-7) = 0
\(\Leftrightarrow\) 4x\(^2\)- 4x + 1 + x\(^2\) + 6x + 9 - 5( x\(^2\) - \(7^2\)) = 0
\(\Leftrightarrow\) 4x\(^2\) - 4x + 1 + x\(^{^{ }2}\) + 6x +9 - 5x\(^2\) + 245 = 0
\(\Leftrightarrow\) 2x + 255=0
\(\Leftrightarrow\) 2x = -255
\(\Leftrightarrow\) x = - \(\frac{255}{2}\)
Vậy x = - \(\frac{255}{2}\)
Chúc bạn học tốt nha ! ^.^
