\(\sqrt{x^2+x-1}-1+\sqrt{x-x^2+1}-1+x-x^2=0\)
\(\Leftrightarrow\dfrac{x^2+x-2}{\sqrt{x^2+x-1}+1}+\dfrac{x-x^2}{\sqrt{x-x^2+1}+1}+x-x^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+2}{\sqrt{x^2+x-1}+1}-\dfrac{x}{\sqrt{x-x^2+1}+1}-x\right)=0\)
\(\Leftrightarrow x=1\)
Áp dụng BĐT: \(\sqrt{ab}\le\dfrac{a+b}{2}\)
Ta có: \(\sqrt{\left(x^2+x-1\right).1}+\sqrt{\left(x-x^2+1\right).1}\)
\(\le\dfrac{x^2+x-1+1}{2}+\dfrac{x-x^2+1+1}{2}=x+1\)\(\Rightarrow\)\(x^2-x+2\le x+1\Leftrightarrow\left(x-1\right)^2\le0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy ...
\(\sqrt{x^2+x-1}+\sqrt{x-x^2+1}=x^2-x+2\)
dk: \(\left\{{}\begin{matrix}x^2+x-1\ge0\\1+x-x^2\ge0\end{matrix}\right.\)\(\Rightarrow\dfrac{\sqrt{5}-1}{2}\le x\le\dfrac{\sqrt{5}+1}{2}\)
có \(\dfrac{\sqrt{5}-1}{2}>0\Rightarrow x>0\)
áp BĐT bunyacoxky c
\(\left(\sqrt{x^2+x-1}\right)^2+\left(\sqrt{x^2+x-1}\right)^2\ge\dfrac{\left(\sqrt{x^2+x-1}+\sqrt{x^2+x-1}\right)^2}{2}\)
\(\Leftrightarrow VT\le\sqrt{4x}=2\sqrt{x}\le x+1\) đẳng thức khi x =1
\(VP=x^2-x+2\ge x+1\)đẳng thức khi x =1
=> \(x=1\) là nghiệm duy nhất
liên hợp hoặc BĐT hoặc đặt ẩn hay cái j c~ dc thiếu j ý tưởng
