Question

[Ly nguyễn gia]

Giải phương trình

1) \(\sqrt{x^2+10x+21}\)+6=3\(\sqrt{x+3}\)+2\(\sqrt{x+7}\)

2) \(\sqrt{x}\)+\(\sqrt{x-5}\)+\(\sqrt{x+7}\)=9

3) (x-2)(x+1)+3(x-2)\(\sqrt{\frac{x+1}{x-2}}\)=10

4) 3x+7\(\sqrt{x-4}\)=14\(\sqrt{x-4}\)-20

5)\(\left(\sqrt{x+1}+1\right)\)\(\left(\sqrt{x+1}+2x-5\right)\)=x

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

Answer 1

[Nguyễn Việt Lâm]

6.

ĐKXĐ: \(x\ge2\)

\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

4.

ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)

\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)

\(\Leftrightarrow3t^2-7t+34=0\)

Phương trình vô nghiệm

5.

ĐKXĐ: ...

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:

\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)

\(\Leftrightarrow2x=4\Rightarrow x=2\)