a: Ta có: \(A=-x^2+2x+5\)

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

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

\(=-\left(x-1\right)^2+6\le6\forall x\)

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=-x^2-8x+10\)

\(=-\left(x^2+8x-10\right)\)

\(=-\left(x^2+8x+16-26\right)\)

\(=-\left(x+4\right)^2+26\le26\forall x\)

Dấu '=' xảy ra khi x=-4

c: Ta có: \(C=-3x^2+12x+8\)

\(=-3\left(x^2-4x-\dfrac{8}{3}\right)\)

\(=-3\left(x^2-4x+4-\dfrac{20}{3}\right)\)

\(=-3\left(x-2\right)^2+20\le20\forall x\)

Dấu '=' xảy ra khi x=2

d: Ta có: \(D=-5x^2+9x-3\)

\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{3}{5}\right)\)

\(=-5\left(x^2-2\cdot x\cdot\dfrac{9}{10}+\dfrac{81}{100}-\dfrac{21}{100}\right)\)

\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{21}{20}\le\dfrac{21}{20}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{9}{10}\)

e: Ta có: \(E=\left(4-x\right)\left(x+6\right)\)

\(=4x+24-x^2-6x\)

\(=-x^2-2x+24\)

\(=-\left(x^2+2x-24\right)\)

\(=-\left(x^2+2x+1-25\right)\)

\(=-\left(x+1\right)^2+25\le25\forall x\)

Dấu '=' xảy ra khi x=-1

f: Ta có: \(F=\left(2x+5\right)\left(4-3x\right)\)

\(=8x-6x^2+20-15x\)

\(=-6x^2-7x+20\)

\(=-6\left(x^2+\dfrac{7}{6}x-\dfrac{10}{3}\right)\)

\(=-6\left(x^2+2\cdot x\cdot\dfrac{7}{12}+\dfrac{49}{144}-\dfrac{529}{144}\right)\)

\(=-6\left(x+\dfrac{7}{12}\right)^2+\dfrac{529}{24}\le\dfrac{529}{24}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{7}{12}\)

tl nhanh giúp mình nha

\(a,\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2021\end{matrix}\right.\\ c,\Leftrightarrow4x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ d,\Leftrightarrow\left(3x+7-x-1\right)\left(3x+7+x+1\right)=0\\ \Leftrightarrow\left(2x+6\right)\left(4x+8\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

Chọn B

\(\Leftrightarrow4\left(x-2\right)+3x\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\)

4(x-2)+3x(x-2)=0

(x-2)(4+3x)=0

x=2 hoặc x=-4/3

`4x-8+3x(x-2)=0`

`=>4(x-2)+3x(x-2)=0`

`=>(x-2)(3x+4)=0`

`=>`$\left[\begin{matrix} x-2=01\\ 3x+4=0\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=2\\ x=-\dfrac{4}{3}\end{matrix}\right.$

Bài 1:

a. $x(x^2-5)=x^3-5x$

b. $3xy(x^2-2x^2y+3)=3x^3y-6x^3y^2+9xy$

c. $(2x-6)(3x+6)=6x^2+12x-18x-36=6x^2-6x-36$

d.

$(x+3y)(x^2-xy)=x^3-x^2y+3x^2y-3xy^2=x^3+2x^2y-3xy^2$

Bài 2:
a.

\((2x+5)(2x-5)=(2x)^2-5^2=4x^2-25\)

b.

\((x-3)^2=x^2-6x+9\)

c.

\((4+3x)^2=9x^2+24x+16\)

d.

\((x-2y)^3=x^3-6x^2y+12xy^2-8y^3\)

e.

\((5x+3y)^3=(5x)^3+3.(5x)^2.3y+3.5x(3y)^2+(3y)^3\)

\(=125x^3+225x^2y+135xy^2+27y^3\)

f.

\((5-x)(25+5x+x^2)=5^3-x^3=125-x^3\)

Bài 3:

a. $x^2+2x=x(x+2)$

b. $x^2-6x+9=x^2-2.3x+3^2=(x-3)^2$
c. $5(x-y)-y(y-x)=5(x-y)+y(x-y)=(x-y)(5+y)$

d. $2x-y^2+2xy-y=(2x-y)+(2xy-y^2)=(2x-y)-y(2x-y)=(2x-y)(1-y)$

e.

$6x^3y^4+12x^2y^3-18x^3y^2=6x^2y^2(xy^2+2y-3x)$

K hiểu 😐😐😐

\(1,\\ a,=x^3-5x\\ b,=3x^3y-6x^3y^2+9xy\\ c,=6x^2-6x-36\\ d,=x^3+2x^2y-3xy^2\\ 2,\\ a,=4x^2-25\\ b,=x^2-6x+9\\ c,=9x^2+24x+16\\ d,=x^3-6x^2y+12xy^2-8y^3\\ e,=125x^3+225x^2y+135xy^2+27y^3\\ f,=125-x^3\)

\(g,=8y^3+x^3\\ 3,\\ a,=x\left(x+2\right)\\ b,=\left(x-3\right)^2\\ c,=\left(x-y\right)\left(y+5\right)\\ d,=2x\left(y+1\right)-y\left(y+1\right)=\left(2x-y\right)\left(y+1\right)\\ e,=6x^2y^2\left(xy^2+2y-3x\right)\)

\(\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\\ =\left(x^3-3x^2+3x-1\right)-\left(x^3+1\right)-\left(1-9x^2\right)\\ =x^3-3x^2+3x-1-x^3-1-1+9x^2\\ =6x^2+3x-3\)

\(=9x^2-6x-x^2+2x-1=8x^2-4x-1\)

3𝑥(3𝑥 − 2) − (𝑥 − 1)^2

=9x2−6x−x2+2x−1

=8x2−4x−1

\(\left(x^3-x^2+3x-3\right):\left(x^2+3\right)\\ =\left[\left(x^3-x^2\right)+\left(3x-3\right)\right]:\left(x^2+3\right)\\ =\left[x^2\left(x-1\right)+3\left(x-1\right)\right]:\left(x^2+3\right)\\ =\left[\left(x^2+3\right)\left(x-1\right)\right]:\left(x^2+3\right)\\ =x-1\)