Question

[Thảo Ngọc]

Giúp e bài 5,6 thui cũng được ạ em cần gấp

Answer 1

[hưng phúc]

Bài 5: 

a. \(\sqrt{x^2+2x+1}=\sqrt{9x^2}\)

<=> \(\sqrt{\left(x+1\right)^2}=\sqrt{\left(3x\right)^2}\)

<=> \(\left|x+1\right|=\left|3x\right|\)

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

<=> \(\left[{}\begin{matrix}x=0,5\\x=-0,25\end{matrix}\right.\)

b. \(\sqrt{x^2-\dfrac{2}{5}x+\dfrac{1}{25}}=\sqrt{\left(2x+1\right)^2}\)

<=> \(\sqrt{\left(x-\dfrac{1}{5}\right)^2}=\sqrt{\left(2x+1\right)^2}\)

<=> \(\left|x-\dfrac{1}{5}\right|=\left|2x+1\right|\)

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

<=> \(\left[{}\begin{matrix}x=-1,2\\x=-\dfrac{4}{15}\end{matrix}\right.\)

c. \(\sqrt{25x^2}=\sqrt{x^4}\)

<=> \(\sqrt{\left(5x\right)^2}=\sqrt{\left(x^2\right)^2}\)

<=> \(\left|5x\right|=\left|x^2\right|\)

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

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

<=> \(\left[{}\begin{matrix}x\left(5-x\right)=0\\x\left(5+x\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\5+x=0\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)