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 \(x=-\dfrac{7}{12}\)

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 😐😐😐

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

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

𝑎)2𝑥−1𝑥−3+4=−1𝑥−3

⇔2x-1x+1x=-3+3-4

⇔2x=-4

⇔x=-2

𝑏)3𝑥−22𝑥+5=6𝑥+14𝑥−3

⇔5+3=6x+14x-3x+22x

⇔8=39x

⇔x=\(\frac{8}{39}\)

𝑐)𝑥+3𝑥+1+𝑥−2𝑥=2

⇔x+3x+x-2x=2-1

⇔3x=1

⇔x=\(\frac{1}{3}\)

𝑑)x+1−2𝑥−3𝑥−1=2𝑥+3𝑥²−1

⇔3x²+2x+2x+3x-x-1-1+1=0

⇔3x²+6x-1=0

⇔3x²+3x+3x+3-4=0

⇔3x(x+1)+3(x+1)-4=0

⇔3(x+1)(x+1)-4=0

⇔3(x+1)²-4=0

⇔(x+1)²=\(\frac{4}{3}\)

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

Vậy ...

a, 2x - x - 3 + 4 = -x - 3

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

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

\(\Leftrightarrow\) 2x = -4

\(\Leftrightarrow\) x = -2

Vậy S = {-2}

b, 3x - 22x + 5 = 6x + 14x - 3

\(\Leftrightarrow\) -19x + 5 = 20x - 3

\(\Leftrightarrow\) -19x - 20x = -3 - 5

\(\Leftrightarrow\) -39x = -8

\(\Leftrightarrow\) x = \(\frac{8}{39}\)

Vậy S = {\(\frac{8}{39}\)}

c, x + 3x + 1 + x - 2x = 2

\(\Leftrightarrow\) 3x + 1 = 2

\(\Leftrightarrow\) 3x = 2 - 1

\(\Leftrightarrow\) 3x = 1

\(\Leftrightarrow\) x = \(\frac{1}{3}\)

Vậy S = {\(\frac{1}{3}\)}

Phần d mình ko hiểu, bạn viết rõ được ko!

Chúc bn học tốt!!

d, x + 1 - 2x - 3x - 1 = 2x + 3x² - 1

\(\Leftrightarrow\) x + 1 - 2x - 3x - 1 - 2x - 3x² + 1 = 0

\(\Leftrightarrow\) -3x² - 6x + 1 = 0

\(\Leftrightarrow\) -(3x² + 6x - 1) = 0

\(\Leftrightarrow\) 3x² + 6x - 1 = 0

\(\Leftrightarrow\) 3x² + 3x + 3x + 3 - 4 = 0

\(\Leftrightarrow\) 3x(x + 1) + 3(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)² - 4 = 0

\(\Leftrightarrow\) (x + 1)² = \(\frac{4}{3}\)

\(\Leftrightarrow\) x + 1 = \(\sqrt{\frac{4}{3}}\) hoặc x + 1 = \(-\sqrt{\frac{4}{3}}\)

\(\Leftrightarrow\) x = \(\sqrt{\frac{4}{3}}\) - 1 và x = \(-\sqrt{\frac{4}{3}}\) - 1

\(\Leftrightarrow\) x = \(\frac{2\sqrt{3}-3}{3}\) và x = \(\frac{-2\sqrt{3}-3}{3}\)

Vậy S = {\(\frac{2\sqrt{3}-3}{3}\); \(\frac{-2\sqrt{3}-3}{3}\)}

Chúc bn học tốt!!

Nguyễn Thị Anh Thư cái này bạn gửi một lần r mà!

5) a) 2x(x^2 - 9) = 0

<=> 2x(x - 3)(x + 3) = 0

<=> x = 0 hoặc x = 3 hoặc x = -3

b) 2x(x - 2021) - x + 2021 = 0

<=> (2x - 1)(x - 2021) = 0

<=> 2x - 1 = 0 hoặc x - 2021 = 0

<=> x = 1/2 hoặc x = 2021

c) 4x^2 - 16x = 0

<=> 4x(x - 4) = 0

<=> x = 0 hoặc x = 4

d) (3x + 7)^2 - (x + 1)^2 = 0

<=> (3x + 7 + x + 1)(3x + 7 - x - 1) = 0

<=> (4x + 8)(2x + 6) = 0

<=> 4x + 8 = 0 hoặc 2x + 6 = 0

<=> x = -2 hoặc x = -3

mình giải tạm nha

2x² - 3x + 10x - 15 - 2x² - 6x - x + 3 = 3

=> Biểu thức trên không phụ thuộc vào biến x.

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

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

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

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

Vì \(\left(x-1\right)^2+2\ge2\) với mọi x

\(\Rightarrow-\left[\left(x-1\right)^2+2\right]\le0\)

hay \(-x^2+2x-3\) luôn âm 

Ta có: -x^2+2x-3

            = - (x^2-2x+3)

            = (x^2 - 2x.1+1+2)

            = -[(x-1)^2 +2]

            Do (x-1)^2 +2 > hoặc = 2 với mọi x

           => -[(x-1)^2+2]< hoặc = 0

           =>-x^2+2x-3 luôn âm

Chúc học tốt !