a) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\) (sửa \(\dfrac{x}{2}\rightarrow x^2\))

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

\(=x^6-\dfrac{1}{27}\)

b) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

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

\(=\dfrac{1}{27}x^3+8y^3\)

Lưu ý : Áp dụng hằng đẳng thức đáng nhớ \(a^3\pm b^3=...\)

Thank you

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

mk c.ơn bn

a, |x^2 - 3x| = 0

=> x^2 - 3x = 0

=> x(x - 3) = 0

=> x = 0 hoặc x - 3 = 0 

=> x = 0 hoặc x = 3

vậy_

\(\left|a^2-3a\right|=0\)

\(\Rightarrow a^2-3a=0\)

\(\Rightarrow a\left(a-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)

Trả lời :

- Theo mk thì nên tặng cô những bức ảnh

chụp cùng hs vì ảnh chứa rất nhiều kỉ niệm

nên sẽ là một món quà ý nghĩa.

- Nếu bn ko cs ảnh thì cs thể tặng cô bó 

hoa, đồ handmade,... Như vậy cx rất ý 

nghĩa mak ! Quan trọng nhất vẫn là tấm lòng !!!

- Lần sau đừng đăng câu hỏi linh tinh lên OLM nha :)))

~Hok tốt~

 Mình bn tặng hay bn vs cả lp tặng

ta co: x⁶=9x⁴

<=>x6:x⁴=9

<=>x²=9

=>x=-3 va 3

\(x^6=9x^4\)

\(x^6\div x^4=9\)

\(x^2=9\)

\(x=3hoặcx=-3\)

Ta có:(3x-7)²⁰⁰⁹=(3x-7)²⁰⁰⁷

\(\Rightarrow\)(3x-7)²⁰⁰⁹-(3x-7)²⁰⁰⁷=0

​\(\Rightarrow\)​(3x-7)²⁰⁰⁷ [(3x-7)²-1]=0​​

\(\Rightarrow\)\(\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Rightarrow\)​\(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)​

\(\Rightarrow\)\(\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\)

​\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=8\end{cases}}\)​

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)

Vậy......

\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)

\(\Rightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)

\(\Rightarrow\left(3x-7\right)^{2007}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x-7=0\\3x-7=\pm1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)