a. \(-x^2+14x-49=0\Leftrightarrow-\left(x-7\right)^2=0\)
\(\Leftrightarrow x-7=0\Leftrightarrow x=7\)
b. \(\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-1+x+2\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=\dfrac{1}{2}\)
c. \(\left(2x-4\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow2\left(x-2\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\Leftrightarrow7x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a)-x^2+14x-49=0
-(x^2-2.x.7+7^2)=0
-(x-7)^2=0
x-7=0
x=7
b)(x-1)^2+2(x-1)(x+2)+(x+2)^2=0
(x-1+x+2)^2=0
(2x+1)^2=0
2x+1=0
2x=-1
x=-1/2
c)(2x-4)(3x+1)+(x-2)^2=0
2(x-2)(3x+1)+(x-2)^2=0
(x-2)(2(3x+1)+x-2)=0
(x-2)7x=0
x-2=0 hoặc 7x=0
x=2 hoặc x=0
Bạn xem lại đề câu d nhe
Nếu đúng đề chắc mình ko bit
\(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)
\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left[\left(x^2-4x\right)-\left(3x-12\right)\right]\left[\left(x^2-x\right)-\left(2x-2\right)\right]=0\)
\(\Leftrightarrow\left[x\left(x-4\right)-3\left(x-4\right)\right]\left[x\left(x-1\right)-2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{1;2;3;4\right\}\)
