Ta có x2−5x+14≐(x−3)2+x+5≥x+5≥x+1+4≥4√x+1x2−5x+14≐(x−3)2+x+5≥x+5≥x+1+4≥4x+1
⇒VT≥VP⇒VT≥VP
Để VT=VP thì x=3.(dấu "=" xảy ra)
ĐK: x + 1 ≥ 0 <=> x ≥ - 1
<=> 16( x + 1) = x4 + 25x2 + 196 - 10x3 + 28x2 - 140x
<=> 16x + 16 = x4 - 10x3 + 53x2 - 140x +196
<=> x4 - 10x3 + 53x2 - 156x + 180 = 0
<=> ( x - 3)2(x2 - 4x + 20 ) = 0
<=> x = 3
\(\left(ĐKXĐ:x\ge-1\right)\)
\(4\sqrt{x-1}-x-5=x^2-6x+9\)
\(\Leftrightarrow\frac{16\left(x+1\right)-\left(x+5\right)^2}{4\sqrt{x+1}+x+5}=\left(x-3\right)^2\)
\(\Leftrightarrow\frac{16x+16-x^2-10x-25}{4\sqrt{x+1}+x+5}-\left(x-3\right)^2\)\(=0\)
\(\Leftrightarrow\frac{-\left(x-3\right)^2}{4\sqrt{x+1}+x+5}-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(-\frac{1}{4\sqrt{x+1}+x+5}-1\right)\)\(=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x=3\)
