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\(BPT\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(3x^2+2x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)
\(\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(x+1\right)\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)
\(\Leftrightarrow\left(x+1\right)\text{[}2+\sqrt{x^2-2x+5}+\frac{2x\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\text{]}\le0\)
\(\Leftrightarrow\left(x+1\right)\left(4\sqrt{x^2+1}+2\sqrt{x^2-2x+5}+2\sqrt{\left(x^2+1\right)\left(x^2-2x+5\right)}+7x^2-4x+5\right)\)\(\le0\Leftrightarrow x+1\le0\Leftrightarrow x\le-1\)
ĐKXĐ: \(-1\le x\le1\)
Xét \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(1+x\right)+\left(1-x\right)+\sqrt{\left(1+x\right)\left(1-x\right)}\right]\)
\(=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)\)
Khi đó phương trình đề trở thành:
\(\sqrt{1+\sqrt{1-x}}\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)=\frac{2+\sqrt{1-x^2}}{3}\)
Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):
\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:
\(\Rightarrow\left(1+\sqrt{1-x^2}\right)\left[\left(1+x\right)+\left(1-x\right)-2\sqrt{\left(1+x\right)\left(1-x\right)}\right]=\frac{1}{3}\)
\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)
Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm
Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)
Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)
\(Xét-mẫu-của-biểu-thức:\left(đk:x\ge1\right).ta-có:x-\sqrt{2\left(x^2+5\right)}=\frac{-\left(x^2+10\right)}{x+\sqrt{2\left(x^2+5\right)}}< 0\\
.\)Vậy nó luôn <0 với đk x>=1
\(Xét-tử:đặt-nó-bằng-A=\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)=2\sqrt{x-1}\left(2x-1\right)-\left(x-1\right)\left(x+4\right)\\ =\sqrt{x-1}\left(2\left(2x-1\right)-\sqrt{x-1\left(x+4\right)}\right)\ge0.\\ \)\(=>\left(2\left(2x-1\right)-\sqrt{\left(x-1\right)}\left(x+4\right)\right)\ge0< =>\frac{\left(5-x\right)\left(x-2\right)^2}{2\left(2x-1\right)+\left(x-1\right)\left(x+4\right)}\ge0< =>x\le5\) Vậy . \(1\le x\le5\)
ĐK: \(x\ge0\)
Với \(x\ge0\Rightarrow\sqrt{\left(x+1\right)^3}-\sqrt{x}>0\)nên bpt \(\Leftrightarrow\sqrt{x\left(x+2\right)}\ge\sqrt{\left(x+1\right)^3}-\sqrt{x}\)
\(\Leftrightarrow x^2+2x\ge x^3+3x^2+4x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\)
\(\Leftrightarrow x^3+2x^2+2x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\le0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2+x+1-2\sqrt{x\left(x+1\right)}\right]\le0\)
\(\Leftrightarrow x^2+x+1-2\sqrt{x\left(x+1\right)}\le0\)
\(\Leftrightarrow\left(\sqrt{x\left(x+1\right)}-1\right)^2\le0\Leftrightarrow\sqrt{x\left(x+1\right)}-1=0\)
\(\Leftrightarrow x=\frac{-1\pm\sqrt{5}}{2}.dox\ge0\Rightarrow x=\frac{-1+\sqrt{5}}{2}\)