Question

[Minh Nguyễn Cao]

Giải phương trình sau

\(\sqrt{8x+1}+\sqrt{46x-10}=-x^3+5x^2+4x+1\)

Answer 1

[ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★]

\(PT\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46x-10}-6=-x^3+5x^2+4x+1-3-6\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)=0\)

Xét \(\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)\)(*) (đk\(\frac{23}{5}\ge x\ge-\frac{1}{8}\))

(*)\(=\frac{8-5\left(\sqrt{8x+1}+3\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}\)

\(=\frac{-7-5\left(\sqrt{8x+1}\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}< 0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy..................

Answer 2

[Tran Le Khanh Linh]

Đề thi thuyển sinh lớp 10 môn Toán Chuyên, TP HCM năm 2012-2013

ĐK \(\frac{-1}{8}\le x\le\frac{23}{5}\)(*) Ta có:

\(\sqrt{8x+1}+\sqrt{46-10x}=-x^3+5x^2+4x+1\)

\(\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46-10x}-6+x^3-x^2-4x^2+4x-8x+8=0\)

\(\Leftrightarrow\frac{8x-1}{\sqrt{8x+1}+3}+\frac{10-10x}{\sqrt{46-10x}+6}+x^2\left(x-1\right)-4x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}+\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8\right)=0\)(**)

(*) \(\Rightarrow-1< x< 5\Rightarrow\left(x+1\right)\left(x+5\right)< 0\Rightarrow x^2-4x-5< 0\)

Và \(\frac{8}{\sqrt{8x+1}+3}< \frac{9}{3}=3\Rightarrow\frac{8}{\sqrt{8x+1}+3}-3< 0\) Do vậy:

\(\frac{8}{\sqrt{8x+1}+3}-\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8< 0\)Do đó:

(**)\(\Leftrightarrow x=1\)

Vậy S={1}