Question

Tìm x:

\(4\sqrt{x^2-x+10-4\sqrt{2-2x}}=x-7+8\sqrt{3-x}\)

 

Answer 1

\(4\sqrt{x^2-x+10-4\sqrt{2-2x}}=x-7+8\sqrt{3-x}=x-7+4\sqrt{\left(3-x\right).4}\le x-7+\dfrac{4\left(3-x+4\right)}{2}=x-7+2\left(3-x+4\right)=7-x\)

\(\Rightarrow4\sqrt{x^2-x+10-4\sqrt{2-2x}}\le7-x\)\(\left(đkxđ:x\le1\right)\)

\(\Leftrightarrow16\left(x^2-x+10\right)-64\sqrt{2-2x}\le\left(7-x\right)^2\)

\(\Leftrightarrow15x^2-2x+111\le64\sqrt{2-2x}\Leftrightarrow\left(x+1\right)^2\left(225x^2-510x+4129\right)\le0\left(1\right)\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\left(tm\right)\)