Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử

a: \(=\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}=\dfrac{31}{35}\)

b: \(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{1}{3}=\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{12+5}{10}=\dfrac{17}{10}\)

c: \(=\dfrac{5\cdot3}{9}=\dfrac{5}{3}\)

d: =5-2/3*6/5

=5-12/15

=5-4/5

=21/5

\(\Rightarrow x\left(9x^2-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{2}{9}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{\sqrt{2}}{3}\\x=-\dfrac{\sqrt{2}}{3}\end{matrix}\right.\)

\(9\cdot3^3\cdot\dfrac{1}{81}\cdot3^2=\dfrac{3^7}{3^4}=27\)

9x3+6=33

k mik mik k lai

9 x 3 + 6

​= 27 + 6

​= 33

​click mk nhé

​

9 x 3 + 6 = 27 + 6 = 33

= 12 nha

k nha linh

=3x3+3=12 nka bn 

27:9x3+3

=3x3+3

=9+3

=12

\(\left\{{}\begin{matrix}x^2+y^2=9\\x^3+y^3=-27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=9\\\left(x+y\right)^3-3xy\left(x+y\right)=-27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b=9\\a^3-3ab=-27\end{matrix}\right.\left(a=x+y;b=xy\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{a^2-9}{2}\\a^3-\dfrac{3a\left(a^2-9\right)}{2}=-27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{a^2-9}{2}\\a^3-27a-54=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{a^2-9}{2}\\\left(a+3\right)^2\left(a-6\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{27}{2}\\a=6\end{matrix}\right.\left(I\right)\) hoặc \(\left\{{}\begin{matrix}b=0\\a=-3\end{matrix}\right.\left(II\right)\)

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}xy=\dfrac{27}{2}\\x+y=6\end{matrix}\right.\Rightarrow\) vô nghiệm

\(\left(II\right)\Leftrightarrow\left\{{}\begin{matrix}xy=0\\x+y=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

\(x^4-9x^3+x^2-9x=0\)

\(\Leftrightarrow x\left(x^2+1\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

\(9+9\times3+18:2\times6\\ =9+9\times3+\left(18:2\right)\times6\\ =9+9\times3+9\times6\\ =9\times\left(1+3+6\right)\\ =9\times10\\ =90\)

_{9 + 9 x 3 + 18 : 2 x 6}

_{= 9 + 9 x 3 + 9 x 6}

_{= 9 x ( 3 + 6 + 1)}

_{= 9 x 10}

_{= 90}