Question

[michelle holder]

gpt a/ \(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)

b/ \(2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}=3-x\)

Answer 1

[Hà Nam Phan Đình]

b) Đặt \(u=\sqrt{1-x}\); \(v=\sqrt{1+x}\)

phương trình trở thành

\(2u-v+3uv=u^2+2\)\(\Rightarrow u^2-2u+v-3uv+2=0\)

lại có \(u^2+v^2=2\)

\(\Rightarrow u^2-2u-3uv+v+u^2+v^2=0\)

\(\Rightarrow\left(u-v-1\right)\left(2u-v\right)=0\)

đến đây thì easy rồi

Answer 2

[Hà Nam Phan Đình]

a)

Đặt \(\sqrt{2x+1}=t\) ;\(\sqrt{x}=k\)

Phương trình trở thành

\(\left(3k^2+t^2\right)t-\left(3t^2+k^2\right)k-1=0\)

\(\Leftrightarrow3k^2t+t^3-3t^2k-k^3-1=0\)

\(\Leftrightarrow\left(t-k\right)\left(t^2+kt+k^2\right)-3tk\left(t-k\right)-1=0\)

\(\Leftrightarrow\left(t-k\right)^3-1=0\)

\(\Leftrightarrow\left(t-k-1\right)\left(\left(t-k\right)^2+t-k+1\right)=0\)

do t > k => t - k > 0

\(\Rightarrow\left(t-k\right)^2+t-k+1>0\)

\(\Rightarrow t-k-1=0\)

\(\Leftrightarrow t=1+k\)\(\Leftrightarrow\sqrt{2x+1}=1+\sqrt{x}\)

\(\Leftrightarrow2x+1=x+2\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

END