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

\(\Leftrightarrow\left(4x^2-4xy+3x\right)+5y^2-y\)

\(\Leftrightarrow\left[4x^2-4x\left(y-\dfrac{3}{4}\right)+\left(y-\dfrac{3}{4}\right)^2\right]+5y^2-y-y^2+\dfrac{3}{2}y-\dfrac{9}{16}\)\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(4y^2-\dfrac{1}{2}y+\dfrac{1}{64}\right)-\dfrac{37}{64}\)

\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(2y-\dfrac{1}{8}\right)^2-\dfrac{37}{64}\ge\dfrac{-37}{64}\)

Vậy Min B = \(\dfrac{-37}{64}\) khi \(\left[{}\begin{matrix}\left(2x-y+\dfrac{3}{4}\right)^2=0\\\left(2y-\dfrac{1}{8}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y-\dfrac{1}{8}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y=\dfrac{1}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{16}+\dfrac{3}{4}=0\\y=\dfrac{1}{16}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11}{32}\\y=\dfrac{1}{16}\end{matrix}\right.\)

\(C=9y^2+2x^2-6y-6xy+5x-1\)

\(=\left(9y^2+6y-6xy\right)+2x^2+5x-1\)

\(=\left[9y^2+6y\left(1-x\right)+\left(1-x\right)^2\right]+2x^2+5x-1-1+2x-x^2\)\(=\left(3y-x+1\right)^2+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{17}{4}\)

\(=\left(3y-x+1\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)

Vậy Min C = \(\dfrac{-17}{4}\) khi \(\left[{}\begin{matrix}\left(3y-x+1\right)^2=0\\\left(x+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-x+1=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-\left(\dfrac{-3}{2}\right)+1=0\\x=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{-5}{6}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

thé này nhé

C=\(x^2+4y^2+1+4xy-4y-2x+x^2-2x+1+5\)

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

đến đây thì  tự đánh giá nhé, tự tim dầu = vậy

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

= x(x - 2y - 3)(x - 2y + 3)

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

= 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

A=(x^2+5x-6)(x^+5x+6)=(x^2+5x)^2-36>=-36

A min=-36 <=> x(x+5)=0

<=>x=0;x=-5

B=(4x^2-4xy+y^2)+(x^2+4x+4)+3=(2x-y)^2+(x+2)^2+3>=3

B min=3 <=> x=-2;y=-4

tick mik nha

Em kiểm tra lại đề câu b.

20x nhe

ho mik lam vs mn