Question

Giải pt

\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)

Answer 1

\(pt\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+48}-7\right)\left(\sqrt{x^2+48}+7\right)}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{\left(\sqrt{x^2+35}-6\right)\left(\sqrt{x^2+35}+6\right)}{\sqrt{x^2+35}+6}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x^2+35}+6}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)

Do : \(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\ne0\)

\(\Rightarrow x=1\)

Answer 2

https://i.imgur.com/r7lKsjx.gif

Answer 3

https://i.imgur.com/zdxkSij.jpg

Answer 4

@Ma Đức Minh@Nguyễn Việt Lâm@Akai Haruma@Phùng Tuệ Minh@Ribi Nkok Ngok@DƯƠNG PHAN KHÁNH DƯƠNG@Mikarin Suzuki