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.

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

tl nhanh giúp mình nha

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

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K hiểu 😐😐😐

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

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

cho mình sửa lại câu d nhé

⇔(x+1)²=\(\frac{4}{3}\)

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

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!!

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

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 \(x=-\dfrac{7}{12}\)

\(A=\left(x+2y\right)^2-2\left(x+y\right)\left(x+2y\right)+\left(x+y\right)^2\\ A=\left(x+2y-x-y\right)^2=y^2\\ A=1000^2=1000000\)

\(\Leftrightarrow2x^2+6x-2x^2=12\Leftrightarrow6x=12\Leftrightarrow x=2\)

\(2x\left(x+3\right)-2x^2=12\\ \Rightarrow2x\left(x+3-x\right)=12\\ \Rightarrow6x=12\\ \Rightarrow x=2\)