Câu hỏi của Nguyễn Thị Khánh

Question

Giải phương trình:

a, $\sqrt{x-7}+\sqrt{x-5}=2$

b, $\sqrt{x^2-6x+9}-3x=2$

c, $\sqrt{3x^2+6x+12}+\sqrt{5x^4+10x^2+9}=3-4x-2x^2$

Nguyễn Việt Lâm · Accepted Answer

a/ ĐKXĐ: ...

$\sqrt{x-7}-\frac{1}{2}+\sqrt{x-5}-\frac{3}{2}=0$

$\Leftrightarrow\frac{x-\frac{29}{4}}{\sqrt{x-7}+\frac{1}{2}}+\frac{x-\frac{29}{4}}{\sqrt{x-5}+\frac{3}{2}}=0$

$\Leftrightarrow\left(x-\frac{29}{4}\right)\left(\frac{1}{\sqrt{x-7}+\frac{1}{2}}+\frac{1}{\sqrt{x-5}+\frac{3}{2}}\right)=0$

$\Leftrightarrow x=\frac{29}{4}$

b/ $\Leftrightarrow\sqrt{x^2-6x+9}=3x+2\left(x\ge-\frac{2}{3}\right)$

$\Leftrightarrow x^2-6x+9=9x^2+12x+4$

$\Leftrightarrow8x^2-18x-5=0\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\x=-\frac{1}{4}\end{matrix}\right.$

c/

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

Do $\left\{{}\begin{matrix}3\left(x+1\right)^2+9\ge9\5x^2\left(x^2+2\right)\ge9\end{matrix}\right.$ $\Rightarrow VT\ge\sqrt{9}+\sqrt{9}=6$

$VP=5-2\left(x+1\right)^2\le5< VP$

Pt luôn vô nghiệmCâu trả lời: a/ ĐKXĐ: ...