a)\(y=x^2-4x+3=\left(x-3\right)\left(x-1\right)\)

b)\(y=x^2+4x+3=\left(x+1\right)\left(x+3\right)\)

c)\(y=-x^2+4x-3=\left(x-3\right)\left(x-1\right)\)

d)\(y=-x^2-4x-3=\left(x+1\right)\left(x+3\right)\)

a) \(\left(x-1\right)^3\)

\(=x^3-3x^2+3x-1\)

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

viết lại đề đi

\(A=25x^2-20x+7\)

\(\Leftrightarrow A=\left(5x-2\right)^2+3\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow5x-2=0\Leftrightarrow x=\frac{2}{5}\)

Vậy \(minA=3\Leftrightarrow x=\frac{2}{5}\)

\(B=-x^2+2x-2\)

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

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

Dấu " = " xảy ra \(\Leftrightarrow x=1\)

Vậy \(maxB=-3\Leftrightarrow x=1\)

\(C=9x^2-12x\)

\(\Leftrightarrow C=\left(9x^2-12x+4\right)-4\)

\(\Leftrightarrow C=\left(3x-2\right)^2-4\ge-4\)

Dấu " = " xảy ra \(\Leftrightarrow3x-2=0\Leftrightarrow x=\frac{2}{3}\)

Vậy \(minC=-4\Leftrightarrow x=\frac{2}{3}\)

\(D=3-10x^2-4xy-4y^2\)

\(\Leftrightarrow D=-\left(4y^2+4xy+x^2+9x^2\right)-3\)

\(\Leftrightarrow D=-\left[\left(2y-x\right)^2+3x^2\right]-3\le-3\)

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

Vậy \(maxD=-3\Leftrightarrow x=y=0\)

\(E=4x-x^2+1\)

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

\(\Leftrightarrow E=-\left(x-2\right)^2+5\le5\)

Dấu " = " xảy ra \(\Leftrightarrow x=2\)

Vậy \(maxE=5\Leftrightarrow x=2\)

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

b: Ta có: \(B=-2x^2+4x+1\)

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

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

\(=-2\left(x-1\right)^2+3\le3\forall x\)

Dấu '=' xảy ra khi x=1

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2