trả lời hộ với mai thi rồi

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

\(\left\{{}\begin{matrix}4x^2+y^2\left(1-4xy\right)=0\\4x^2+2y^2-4xy-1=0\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4xy=-1\\y^2=1\end{matrix}\right.\)

Bạn tự giải nốt

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

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

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

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

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

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

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

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

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

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

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

Đề bài có đúng không bạn?

đề thiếu y³ nha

x^3-4x^2y+4xy^2-y³

=(x³-y³)-(4x²y-4xy²)

=(x-y)(x²+xy+y²)-4xy(x-y)

=(x-y)(x²-3xy+y²)

đề bài cô cho như vậy
 

Đề bài yêu cầu gì?

tính giá trị nhỏ nhất của biểu thức:A=x²+10y²​+4xy-4x-2y+20

Ta có : A=x²+10y²​+4xy-4x-2y+20

→ A = \(\left(x^2+4xy+4y^2\right)-\left(4x+8y\right)+4+\left(6y^2+6y+\dfrac{6}{4}\right)-\dfrac{6}{4}+16\)

→ A = \(\left[\left(x+2y\right)^2-4\left(x+2y\right)+2^2\right]+6\left(y^2+\dfrac{2.y}{2}+\dfrac{1}{4}\right)-\dfrac{6}{4}+16\)

→ A = \(\left(x+2y-2\right)^2+6\left(y+\dfrac{1}{2}\right)^2+\dfrac{29}{2}\) ≥ \(\dfrac{29}{2}\)

Vì \(\left(x+2y-2\right)^2\ge0\)    ∀ x,y

\(6\left(y+\dfrac{1}{2}\right)^2\ge0\)     ∀ y

⇒ GTNN của A là \(\dfrac{29}{2}\)

Dấu = xảy ra ⇔ \(\left\{{}\begin{matrix}\left(x+2y-2\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2y-2=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy A đạt GTNN là \(\dfrac{29}{2}\)  \(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

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)

lai hoi bo kien thuc rong ak

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

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

Do \(\left\{{}\begin{matrix}-\left(2x-y\right)^2\le0\\-\left(y-1\right)^2\le0\end{matrix}\right.\) ;\(\forall x;y\)

\(\Rightarrow A\le4;\forall x;y\)

Vậy \(A_{max}=4\) khi \(x=\dfrac{1}{2};y=1\)