`a, x^2 - 3x + 5 = x^2 - 2 . 3/2. x + 9/4 + 7/4 = (x-3/2)^2 + 7/4 >= 0 + 7/4 = 7/4`.
Dấu bằng xảy ra `<=> x = 3/2`.
`b, 2(x^2 - 5x + 1) = 2(x^2 - 2 . 5/2x + 25/4 - 21/4) = 2(x-5/2)^2 - 21/2 >= -21/2`
Dấu bằng xảy ra `<=> x = 5/2`.
`e, = x^2 - 10x + 25 + 4x^2 + 4xy + y^2 - 35 = (x-5)^2 + (2x+y)^2 - 35 >= 0 - 35 = -35`
Dấu bằng xảy ra `<=> x = 5 -> y = -10`
`f,` Đặt `t = x^2 - 7x`.
`-> t(t +12) = t^2 + 12t + 36 - 36 >= (t+6)^2 - 36 >= -36`
Dấu bằng xảy ra `<=> t + 6 = 0 -> x^2 - 7x + 6 = 0 -> x = 1, 6`.
`a.`\(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
Dấu "=" `<=>x=3/2`
`b.`\(A=2x^2-10x+1=2\left(x^2-5x+\dfrac{1}{2}\right)=2\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{23}{2}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{23}{2}\ge-\dfrac{23}{2}\)Dấu "=" `<=>x=5/2`
`c.`\(C=\left(2x-1\right)^2+\left(x+2\right)^2=4x^2-4x+1+x^2+4x+4=5x^2+5\ge5\)
Dấu "=" `<=>x=0`
`d.`\(\left(1-3x\right)^2+\left(5y+2\right)^4-1\ge-1\)
Dấu "=" `<=>` \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
`e.`\(E=5x^2-10x+4xy+y^2-10=\left(4x^2+4xy+y^2\right)+\left(x^2-10x+25\right)-35=\left(2x+y\right)^2+\left(x-5\right)^2-35\ge-35\)
Dấu "=" `<=>`\(\left\{{}\begin{matrix}x=5\\y=-10\end{matrix}\right.\)
`f.`\(F=\left(x^2-7x\right)\left(x^2-7x+12\right)\)
Đặt \(x^2-7x=t\)
`=>`\(F=t\left(t+12\right)\)
\(F=t^2+12t=\left(t^2+12t+36\right)-36=\left(t+6\right)^2-36\ge-36\)
Dấu "=" xảy ra \(\Leftrightarrow t=-6\)
\(\Rightarrow x^2-7x=-6\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
`c, 4x^2 - 4x + 1 + x^2 + 4x + 4`
`= 5x^2 + 5 >= 0 + 5 = 5`.
Dấu bằng xảy ra `<=> x = 0`.
`d, (1-3x)^2 >= 0 forall x in RR`.
Dấu `=` xảy ra `<=> x = 1/3`.
`(5y+2)^4 >= forall y in RR`
Dấu bằng xảy ra `<=> y = -2/5.`
`-> D >= 0 + 0 - 1 >= -1`
