a, Ta có : \(-x^2+2x-1-3\)

\(=-\left(x-1\right)^2-3\)

Ta thấy : \(\left(x-1\right)^2\ge0\forall x\)

=> \(-\left(x-1\right)^2-3\le-3\forall x\)

Vậy Max = -3 <=> x = 1 .

b, Ta có : \(-x^2-4x-4+4\)

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

Ta thấy : \(\left(x+2\right)^2\ge0\forall x\)

=> \(-\left(x+2\right)^2+4\le4\forall x\)

Vậy Max = 4 <=> x = -2 .

c, Ta có : \(-9x^2+24x-16-2\)

\(=-9\left(x^2-\frac{2.4x}{3}+\frac{16}{9}\right)-2\)

\(=-9\left(x-\frac{4}{3}\right)^2-2\)

Ta thấy : \(\left(x-\frac{4}{3}\right)^2\ge0\forall x\)

=> \(-9\left(x-\frac{4}{3}\right)^2-2\le-2\forall x\)

Vậy Max = -2 <=> x = \(\frac{4}{3}\) .

d, Ta có : \(-x^2+4x-4+3\)

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

Ta thấy : \(\left(x-2\right)^2\ge0\forall x\)

=> \(-\left(x-2\right)^2+3\le3\forall x\)

Vậy Max = 3 <=> x = 2 .

e, Ta có : \(-x^2+2x-1-4y^2-4y-1+7\)

\(=-\left(x-1\right)^2-4\left(y^2+y+\frac{1}{4}\right)+7\)

\(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\)

Ta thấy : \(\left\{{}\begin{matrix}\left(x-1\right)^2\\\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\ge0\forall xy\)

=> \(\left\{{}\begin{matrix}-\left(x-1\right)^2\\-4\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\le0\forall xy\)

=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2\le0\forall xy\)

=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\le7\forall xy\)

Vậy Max = 7 <=> \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)

$x(x+y)+4x+4y$

$=x(x+y)+4(x+y)$

$=(x+y)(x+4)$

a) \(4x^2-12x+9\)

\(=\left(2x\right)^2-2.2.3+3^2\)

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

b) \(4x^2+4x+1\)

\(=\left(2x\right)^2+2.2x.1+1^2\)

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

c) \(1+12x+36x^2\)

\(=1^2+2.6x+\left(6x\right)^2\)

\(=\left(1+6x\right)^2\)

d) \(9x^2-24xy+16y^2\)

\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)

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

e) Viết = công thức trực quan hộ mình

f) \(-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x^2-2.5x+5^2\right)\)

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

=>x^2-4x+2y^2-4y+6=0

=>x^2-4x+4+2y^2-4y+2=0

=>(x-2)^2+2(y-1)^2=0

=>x=2 và y=1

Ta có: 4x² - y² + 4x + 4y - 3

= (4x² - 4x + 1) - (y² - 4y + 4)

= (2x - 1)² - (y - 2)²

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

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

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

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

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

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

\(=\left(2x+y-1\right)\left(2x-y+3\right)\)

ax - by + 4x - 4y

= (ax + 4x) - (by + 4y)

= x(a + 4) - y(b + 4)

\(\Leftrightarrow x^3-ax^2-4=\left(x^2+4x+4\right)\cdot a\left(x\right)=\left(x+2\right)^2\cdot a\left(x\right)\)

Thay \(x=-2\Leftrightarrow-8-4a-4=0\Leftrightarrow a=-3\)

\(\Leftrightarrow x^3+4x^2+4x+\left(-4-a\right)x^2-4⋮x^2+4x+4\)

\(\Leftrightarrow-4-a=4+x^2\)

\(\Leftrightarrow a=-4-4-x^2=-x^2-8\)

a.

\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)

Đặt \(x^2-x+1=y\) ta được:

\(y\left(y+1\right)=12\)

\(\Leftrightarrow y^2+y-12=0\)

\(\Leftrightarrow y^2+4y-3y-12=0\)

\(\Leftrightarrow\left(y-3\right)\left(y+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=3\\y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=3\\x^2-x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b.

\(3y^3-7y^2-7y+3=0\)

\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y-1\right)\left(y-3\right)=0\)

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

TÌM GTNN , MIN ẤY