a, Ta có: \(x^2\ge0\)
\(\Leftrightarrow x^2+5\ge5\)
\(\Leftrightarrow\left(x^2+5\right)^2\ge25\)
\(\Leftrightarrow\left(x^2+5\right)^2+4\ge29\)
Dấu " = " khi \(x^2=0\Leftrightarrow x=0\)
Vậy \(MIN_{\left(x^2+5\right)^2+4}=29\) khi x = 0
c, Đặt \(C=x\left(x-6\right)+100\)
\(=x^2-6x+100=x^2-6x+9+91\)
\(=\left(x-3\right)^2+91\)
Ta có: \(\left(x-2\right)^2+91\ge91\)
Dấu " = " khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy \(MIN_C=91\) khi x = 2
b,
Q = \(4x^2+2x-5\)
\(=4\left(x^2+\dfrac{1}{2}x-\dfrac{5}{4}\right)\)
\(=4\left(x^2+2.x.\dfrac{1}{4}+\dfrac{1}{16}\right)-5-\dfrac{1}{4}\)
\(=4\left(x+\dfrac{1}{4}\right)^2-\dfrac{21}{4}\)
Mà \(4\left(x+\dfrac{1}{4}\right)^2\ge0=>4\left(x+\dfrac{1}{4}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)
Vậy \(Min_Q=-\dfrac{21}{4}\Leftrightarrow x=-\dfrac{1}{4}\)
Vì tú ko lm câu b nên mk chỉ làm câu b thoy .
a) (x2+5)2+4 = x4+10x2+25+4=x4+10x2+29
b)4x2+2x-5=Em ko biết 8==(==D) Phê quá
