Question

a) \(\left(x^2-5x+6\right)\left(\sqrt{x+5}+4\right)=3x^3-10x^2-7x+30\)

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

c) \(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)

Answer 1

a.

ĐKXĐ: \(x\ge-5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+6=0\\\sqrt{x+5}+4=3x+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\\sqrt{x+5}=3x+1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\x+5=9x^2+6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\9x^2+5x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{4}{9}\end{matrix}\right.\)

Answer 2

b. Bạn coi lại đề, pt này nghiệm rất xấu

c.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)