a,A=\(-5x^2+10x-7=-2-5\left(x^2-2x+1\right)=-2-5\left(x-1\right)^2\)
Có \(-5\left(x-1\right)^2\le0\) với mọi x
<=> \(-2-5\left(x-1\right)^2\le-2\) vs mọi x
<=> \(A\le-2\)
Dấu "=" xảy ra <=> x=1
Vậy maxA=-2 <=> x=1
b,B=\(-5x^2-4x+1=1+\frac{4}{5}-5\left(x^2+2.\frac{4}{10}x+\frac{4}{25}\right)\)
=1+\(\frac{4}{5}-5\left(x+\frac{4}{10}\right)^2\)
vì \(-5\left(x+\frac{4}{10}\right)^2\le0\) vs mọi x
<=> \(1+\frac{4}{5}-5\left(x+\frac{4}{10}\right)^2\le1+\frac{4}{5}\)
<=> B\(\le1+\frac{4}{5}\)
Dấu "=" xảy ra<=> x=-\(\frac{4}{10}=-\frac{2}{5}\)
Vậy maxB=\(\frac{9}{5}\) <=>x \(=-\frac{2}{3}\)
c,C=\(\frac{3}{4x^2-4x+5}=\frac{3}{\left(2x-1\right)^2+4}\)
Có \(\left(2x-1\right)^2+4\ge4\) vs mọi x
<=> \(\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\) vs mọi x
<=> \(C\le\frac{3}{4}\)
Dấu "=" xảy ra<=> x=\(\frac{1}{2}\)
Vậy maxC=\(\frac{3}{4}\) <=> \(x=\frac{1}{2}\)
