Question

Bài 1 : Tìm x,y,z

a) \(\sqrt{\frac{5x+7}{x+3}}\)=4

b) x+y+13=\(4\sqrt{x}\)+6√y-1

c) \(3\sqrt{2}\)=√x-√2x-1

d) √x^2+6x+9 =3

Mn giúp mình với ạ

Answer 1

Bài 1:

a) \(\sqrt{\frac{5x+7}{x+3}}=4\)

\(\Leftrightarrow\frac{5x+7}{x+3}=16\)

\(\Leftrightarrow5x+7=16x+48\)

\(\Leftrightarrow x=\frac{-41}{11}\)

b) \(x+y+13=4\sqrt{x}+6\sqrt{y}-1\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=-1\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=-1\)( vô lí )

Vậy pt vô nghiệm

c) Không hiểu đề như nào ??

d) \(\sqrt{x^2+6x+9}=3\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3\)

\(\Leftrightarrow\left|x+3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=3\\x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)