a) \(x^2-25-4xy+4y^2\)

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

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

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

b) \(x^2-8x+15\)

\(=x^2-3x-5x+15\)

\(=x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x-5\right)\)

a)\(x^2-25-4xy+4y^2\Leftrightarrow\left(x^2-4xy+4y^2\right)-25\)

\(\Leftrightarrow\left(x-2y\right)^2-5^2\)

\(\Leftrightarrow\left(x-2y-5\right)\left(x-2y+5\right)\)

b)\(x^2-8x+15\Leftrightarrow\left(x-3\right)\left(x-5\right)\)

a) \(A=x^2-6x+25\)

\(=\left(x^2-6x\right)+25\)

\(=\left(x^2-6x+3^2\right)+16\)

\(=\left(x-3\right)^2+16\)

Ta có \(\left(x-3\right)^2\ge0\\ \Rightarrow\left(x-3\right)^2+16\ge16\)

Dấu ''='' xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy GTNT của A là 16 khi x = 3

a) \(A=x^2-6x+25\)

\(A=x^2-2.x.3+9-9+25\)

\(A=\left(x-3\right)^2+16\)

Vì \(\left(x-3\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

\(\Rightarrow Amin=16\Leftrightarrow x-3=0\Rightarrow x=3\)

Vậy Amin = 16 <=> x = 3

b) \(B=5x^2-4x+3\)

\(B=5\left(x^2-\dfrac{4}{5}x+\dfrac{3}{5}\right)\)

\(B=5\left(x^2-2.x.\dfrac{2}{5}+\dfrac{4}{25}-\dfrac{4}{25}+\dfrac{3}{5}\right)\)

\(B=5\left(x^2-2.x.\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{11}{25}\right)\)

\(B=5\left(x-\dfrac{2}{5}\right)^2+\dfrac{11}{5}\)

Vì \(5\left(x-\dfrac{2}{5}\right)^2\ge0\) với mọi x

\(\Rightarrow5\left(x-\dfrac{2}{5}\right)^2+\dfrac{11}{5}\ge\dfrac{11}{5}\)

\(\Rightarrow Bmin=\dfrac{11}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Rightarrow x=\dfrac{2}{5}\)

Vậy Bmin = 11/5 <=> x = 2/5

c) \(C=x^2-4xy+5y^2-4y+13\)

\(C=x^2-2.x.2y+\left(2y\right)^2+y^2-2.y.2+4+9\)

\(C=\left(x-2y\right)^2+\left(y-2\right)^2+9\)

Vì \(\left(x-2y\right)^2+\left(y-2\right)^2\ge0\) với mọi x và y

\(\Rightarrow\left(x-2y\right)^2+\left(y-2\right)^2+9\ge9\)

\(\Rightarrow Cmin=9\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

Vậy Cmin = 9 <=> x = 4 và y = 2

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

a/ \(4x^2+2y^2-4xy+4x-2y+5=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+2\left(2x-y\right)+1+4=0\)

\(\Leftrightarrow\left(2x-y\right)^2+2\left(2x-y\right)+1+4=0\)

\(\Leftrightarrow\left(2x-y+1\right)^2+4=0\)

Với mọi x, y ta có :

\(\left(2x-y+1\right)^2\ge0\Leftrightarrow\left(2x-y+1\right)^2+4>0\)

\(\Leftrightarrow pt\) vô nghiệm

\(25-x^2+4xy-4y^2\)

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

\(=25-\left(x-2y\right)^2\)

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

Vậy..

giúp câu này luôn với :)))

Tìm GTLN của \(P=-x^2+3x+4\)

\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)

\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)

\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)

WTF đăng một loạt vầy ai dám làm @@

Mấy bài này trong sách bài tập cx có bài mẫu

tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết

\(A=x^2+2x+3=\left(x+1\right)^2+2>=2\)

Dấu '=' xảy ra khi x=-1

\(B=-\left(x^2+4x-1\right)\)

\(=-\left(x^2+4x+4-5\right)\)

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

Dấu '=' xảy ra khi x=-2

\(C=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21< =21\)

Dấu '=' xảy ra khi x=-4

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

\(=-\left(x^2+x+\dfrac{1}{4}-\dfrac{5}{4}\right)\)

\(=-\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}< =\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=-1/2