X nhân X = 27 nhân 3
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\(\frac{3}{4}.\frac{12}{7}+x=\frac{5}{2}\)
\(\frac{3.3}{1.7}+x=\frac{5}{2}\)
\(\frac{9}{7}+x=\frac{5}{2}\)
\(x=\frac{5}{2}-\frac{9}{7}\)
\(x=\frac{35}{14}-\frac{18}{14}\)
\(x=\frac{17}{14}\)
\(\left(x+25\right)\left(x^3-27\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+25=0\\x^3-27=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-25\\x^3=3^3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-25\\x=3\end{cases}}\)
Vậy : \(x\in\left\{-25,3\right\}\)
\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu
\(\left(\dfrac{9}{16}\right)^5\cdot x=\left(\dfrac{27}{64}\right)^3\)
\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{10}\cdot x=\left(\dfrac{3}{4}\right)^9\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^9:\left(\dfrac{3}{4}\right)^{10}\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^{-1}\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy x=4/3
(\(\dfrac{9}{16}\))5\(\times\) \(x\) = (\(\dfrac{27}{64}\))3
\(x\) = (\(\dfrac{27}{64}\))3 : (\(\dfrac{9}{16}\))5
\(x\) = (\(\dfrac{3^3}{2^6}\))3: (\(\dfrac{3^2}{2^4}\))5
\(x\) = \(\dfrac{3^9}{2^{18}}\) : \(\dfrac{3^{10}}{2^{20}}\)
\(x\) = \(\dfrac{3^9}{2^{18}}\) \(\times\) \(\dfrac{2^{20}}{3^{10}}\)
\(x\) = \(\dfrac{2^2}{3}\)
\(x\) = \(\dfrac{4}{3}\)
X x X=81
X x X=9x9
X=9
XxX=27x3
XxX=81
XxX=9x9
X=9