Question

Chi tiết giúp tớ ạ !

Answer 1

Câu 5:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x+4>=0\\-x^2+4x+14>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-4\\x^2-4x-14< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-4\\x^2-4x+4< =18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x-2\right)^2< =18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-4\\-3\sqrt{2}< =x-2< =3\sqrt{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-4\\-3\sqrt{2}+2< =x< =3\sqrt{2}+2\end{matrix}\right.\Leftrightarrow-3\sqrt{2}+2< =x< =3\sqrt{2}+2\)

=>Sai

b: \(\left(x+1\right)\left(\sqrt{x+4}-\sqrt{-x^2+4x+14}\right)=0\)

=>\(\left[{}\begin{matrix}x+1=0\\\sqrt{x+4}-\sqrt{-x^2+4x+14}=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\left(nhận\right)\\x+4=-x^2+4x+14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+4+x^2-4x-14=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\\x^2-3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x-5\right)\left(x+2\right)=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=5\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

=>Phương trình có 3 nghiệm phân biệt

=>Đúng

c: Sai

d: Tổng các nghiệm là:

-1+5+(-2)=5-3=2

=>Đúng

Câu 6: 

a: ĐKXĐ: \(\left\{{}\begin{matrix}x+3>=0\\x^2+2x-3>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-3\\\left(x+3\right)\left(x-1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-3\\\left[{}\begin{matrix}x>=1\\x< =-3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=1\\x=-3\end{matrix}\right.\)

=>Sai

b: \(\left(\sqrt{x^2+2x-3}-2x+2\right)^2+\left(2-\sqrt{x+3}\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x^2+2x-3}-2x+2=0\\2-\sqrt{x+3}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x^2+2x-3}=2x-2\\\sqrt{x+3}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2+2x-3=\left(2x-2\right)^2\\2x-2>=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+3\right)\left(x-1\right)-4\left(x-1\right)^2=0\\x>=1\\x=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+3-4x+4\right)=0\\x>=1\\x=1\end{matrix}\right.\Leftrightarrow x=1\)

=>Phương trình có nghiệm duy nhất là x=1

=>Sai

c: Sai

d: Đúng