Question

[Trần Hoàng Đạt]

Giải phương trình vô tỉ:

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

b) \(\sqrt{3x+8+6\sqrt{3x-1}}+\sqrt{3x+8-6\sqrt{3x-1}}=3x+4\)

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

Answer 1

[Akai Haruma]

a) ĐKXĐ: \(-1\leq x\leq 2\)

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

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

Đặt \(\sqrt{2+x-x^2}=t(t\geq 0)\). PT trở thành:

\(t=-3+2t^2\)

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

\(\Rightarrow t=\frac{3}{2}\) (do \(t\geq 0)\)

\(\Rightarrow 2+x-x^2=\frac{9}{4}\Rightarrow x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow (x-\frac{1}{2})^2=0\Rightarrow x=\frac{1}{2}\) (thỏa mãn)

Answer 2

[Akai Haruma]

b) ĐK: \(x\geq \frac{1}{3}\)

PT \(\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4\)

\(\Leftrightarrow \sqrt{(\sqrt{3x-1}+3)^2}+\sqrt{(\sqrt{3x-1}-3)^2}=3x+4\)

\(\Leftrightarrow \sqrt{3x-1}+3+|\sqrt{3x-1}-3|=3x+4\)

\(\Leftrightarrow |\sqrt{3x-1}-3|=3x-\sqrt{3x-1}+1\)

Nếu \(\sqrt{3x-1}\geq 3\):

\(\Rightarrow \sqrt{3x-1}-3=3x-\sqrt{3x-1}+1\)

\(\Leftrightarrow 3x+4-2\sqrt{3x-1}=0\)

\(\Leftrightarrow (3x-1)-2\sqrt{3x-1}+5=0\)

\(\Leftrightarrow (\sqrt{3x-1}-1)^2+4=0\) (vô lý)

Nếu \(\sqrt{3x-1}< 3\):

\(\Rightarrow 3-\sqrt{3x-1}=3x-\sqrt{3x-1}+1\)

\(\Leftrightarrow 3x=2\Rightarrow x=\frac{2}{3}\) (thỏa mãn)

Vậy...........