Question

[Kathy Nguyễn]

Giai phuong trinh

1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)

2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)

3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)

4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)

5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)

6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)

7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)

Answer 1

[Nguyễn Thị Ngọc Thơ]

Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)

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

Vậy ....