Ta có: \(a^4+1\ge a\left(a^2+1\right)\)\(\Leftrightarrow a^4+1\ge a^3+a\)

\(\Leftrightarrow a^4-a^3+1-a\ge0\)

\(\Leftrightarrow a^3\left(a-1\right)-\left(a-1\right)\ge0\)

\(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(a^2+a+1\right)\ge0\)

mà \(a^2+a+1=a^2+2a\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}\)\(=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(\frac{3}{4}>0\right)\)

Vì \(\left(a+\frac{1}{2}\right)^2\ge0\)với mọi a ( Đó là điều hiển nhiên )

Vậy...................

Bài làm chỉ mang tính chất tượng trưng còn sai sót

bình phương được

\(10+2\sqrt{10x-x^2}=16\)

\(\Leftrightarrow\sqrt{10x-x^2}=3\Leftrightarrow10x-x^2=9\Leftrightarrow x=5\pm\sqrt{34}\)

\(\sqrt{x}+\sqrt{10-x}=4\) (ĐKXĐ: \(0< x< 10\))

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{10-x}\right)^2=4^2\)

\(\Leftrightarrow x+10-x+2\sqrt{x\left(10-x\right)}=16\)

\(\Leftrightarrow10+2\sqrt{x\left(10-x\right)}=16\)

\(\Leftrightarrow2\sqrt{x\left(10-x\right)}=6\)

\(\Leftrightarrow\sqrt{x\left(10-x\right)}=3\)

\(\Leftrightarrow x\left(10-x\right)=9\)

\(\Leftrightarrow10x-x^2=9\)

\(\Leftrightarrow-x^2+10x-9=0\)

\(\Leftrightarrow x^2-10+9=0\)(nhân cả 2 vế với -1)

\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

Vậy nghiệm của PT là \(x\in\left\{1;9\right\}\)

ai giúp mk vs

\(VT=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right).\left(15+2\sqrt{6}\right)=201\)

     \(=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right).\left(15+2\sqrt{6}\right)\)

    \(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)\)

   \(=15^2-\left(2\sqrt{6}\right)^2=201=VP\)   (đpcm)