ĐKXĐ: x lớn hơn hoặc bằng -1 và x nhỏ hơn hoặc bằng 1.
\(4+2\sqrt{1-x}=-3x+5\sqrt{x+1}+\sqrt{1-x^2}\)
\(\Leftrightarrow4+2\left(\sqrt{1-x}-1\right)+2=-3x+5\left(\sqrt{x+1}-1\right)+\left(\sqrt{1-x^2}-1\right)+5+1\)
\(\frac{-2x}{\sqrt{1-x}+1}=-3x+\frac{5x}{\sqrt{x+1}+1}-\frac{x^2}{\sqrt{1-x^2}+1}\Leftrightarrow x\left(\frac{x}{\sqrt{1-x^2}+1}-\frac{5}{\sqrt{x+1}+1}-\frac{2}{\sqrt{1-x}+1}+3\right)=0\)
\(\Leftrightarrow x=0.\)
\(Pt\Leftrightarrow3\left(x+1\right)+2\sqrt{1-x}+1=5\sqrt{x+1}+\sqrt{1-x^2}\)
đặt \(\sqrt{x+1}=a,\sqrt{1-x}=b\)
\(\Leftrightarrow3a^2+2b+1=a\left(5+b\right)\)
\(\Leftrightarrow3a^2-\left(5+b\right)a+2b+1=0\)
\(\Delta=b^2-4ac=\left(-b-5\right)^2-4.3.\left(2b+1\right)\)
\(=b^2+10b+25-24b-12\)
\(=b^2-14b+13\)
\(TH1:\Rightarrow a=\frac{5+b+\sqrt{b^2-14b+13}}{6}\)
\(\Rightarrow6a-5-b=\sqrt{b^2-14b+13}\)
\(\Rightarrow6\sqrt{1+x}-5-\sqrt{1-x}=\sqrt{1-x-14\sqrt{1-x}+13}\)
\(\hept{\begin{cases}x=0\left(nhan\right)\\x=......\left(loai\right)\end{cases}}\)
TH2:\(a=\frac{5+b-\sqrt{b^2-14b+13}}{6}\)
\(.............................................\)
cách này hơi dài.
Ta có \(\left(a+b+c+1\right)^2=\left(\left(a+1\right)+b+c\right)^2\)
\(=a^2+2a+1+b^2+c^2+2b+2c+2ab+2bc+2ac\left(f\right)\)
Từ \(\left(f\right)\Rightarrow2a+2b+2c+2ab+2bc+2ac\ge3\left(a^2+b^2+c^2\right)\)
Mà \(a,b,c\in\left[0;1\right]\)nên \(a\ge a^2,b\ge b^2,c\ge c^2\left(g\right)\)
Từ \(\left(f\right)vs\left(g\right)\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\)
Áp dụng bất đẳng thức cô si ta có:
\(2ab+2bc+2ac\ge3\sqrt[3]{8\left(abc\right)^2}=6\sqrt[3]{\left(abc\right)^2}\)
\(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}\)
\(\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\Rightarrowđpcm\).Dấu bằng tự tìm nha.
