\(x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(=\left(2y-x-5\right)^2+\left(y-1\right)^2+2\)

Ta có :

\(\left(2y-x-5\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2y-x-5\right)^2+\left(y-1\right)^2+2\ge2\forall x\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=0\\\left(2y-x-5\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=0\\2y-x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-3\end{matrix}\right.\)

Vậy biểu thức đạt GTNN = 2 ⇔ \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

a) ta có : \(N=-21x^{99}-21x^{98}-...-21x^2-21x\)

\(\Rightarrow xN=-21x^{100}-21x^{99}-...-21x^2-21x^2\)

\(\Rightarrow xN-N=-21x^{100}+21x\)

\(\Leftrightarrow\left(x-1\right)N=-21x^{100}+21x\Leftrightarrow N=\dfrac{21x-21x^{100}}{x-1}\)

\(\Rightarrow A=x^{100}-21x^{99}-21x^{98}-...-21x^2-21x+2010\)

\(=x^{100}+\dfrac{21x-21x^{100}}{x-1}+2010\)

\(=\dfrac{21x-21x^{100}+x^{101}-x^{100}+2010x-2010}{x-1}\)

\(=\dfrac{x^{101}-22x^{100}+2031x-2010}{x-1}\)

thay \(x=22\) ta có : \(A=\dfrac{22^{101}-22.22^{100}+2031.22-2010}{22-1}\)

\(=\dfrac{22^{101}-22^{101}+2031.22-2010}{21}=\dfrac{2031.22-2010}{21}=2032\)

vậy ............................................................................................................

câu b lm tương tự .

y^3 - y^2 - 21y + 45 = 0

<=> y^3 - 3y^2 + 2y^2 - 6y - 15y + 45 = 0

<=> y^2(y - 3) + 2y(y - 3) - 15(y - 3) = 0

<=> (y^2 + 2y - 15)(y-3) = 0

<=> (y^2 + 5y - 3y - 15)(y - 3) = 0

<=> [y(y+5) - 3(y-5)](y-3) = 0

<=> (y-3)(y+5)(y-3) = 0

<=> y- 3 = 0 hoặc y + 5 = 0

<=> y = 3 hoặc y = -5

GTNN nak !!!

\(B=x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+\left(y^2-2y+1\right)+27\)

\(=\left[\left(x-2y\right)^2+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\) có GTNN là 2

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy \(B_{min}=2\) tại \(x=-3;y=1\)

= x^2-4xy+4y^2+y^2-22y+121-93

=(x+2y)^2+(y-11)^2>=-93

GNNN là -93

Ta có: \(B=x^2-4xy+5y^2-22y+28\)

                \(=x^2-4xy+y^2-22y+121-93\)

                  \(=\left(x-2y\right)^2+\left(y-11\right)^2-93\)

Vì \(\left(x-2y\right)^2\ge0;\left(y-11\right)^2\ge0\)

\(\Rightarrow B\ge-93\)

Dấu "=" xảy ra khi \(y-11=0\Rightarrow y=11\)

                              \(x-2y=0\Rightarrow x-2.11=0\Rightarrow x=22\)

Vậy B_min=-93 khi x=22; y=11