15 con ga

9 con lon 

tick cho mih xong mih cho loi giai

\(\left\{{}\begin{matrix}ab=2\\bc=3\\ac=54\end{matrix}\right.\Rightarrow\left(abc\right)^2=2\cdot3\cdot54=324\)

\(\Rightarrow\left[{}\begin{matrix}abc=\sqrt{324}\\abc=-\sqrt{324}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}abc=18\\abc=-18\end{matrix}\right.\)

Nếu abc = 18

\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{18}{2}=9\\a=\dfrac{18}{3}=6\\b=\dfrac{18}{54}=\dfrac{1}{3}\end{matrix}\right.\)

Nếu abc = -18

\(\Rightarrow\left\{{}\begin{matrix}c=-\dfrac{18}{2}=-9\\a=-\dfrac{18}{3}=-6\\b=-\dfrac{18}{54}=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(\left(a,b,c\right)=\left(-6;-\dfrac{1}{3};-9\right);\left(6;\dfrac{1}{3};9\right)\)

\(y\in N\) => 3^y lẻ

Mà 242 chẵn => 2^x lẻ

=> x = 0

=> 2⁰ + 242 = 3^y

<=> 1 + 242 = 3^y

<=> 243 = 3^y <=> y = 5

Gọi ba phần đó là a,b,c

=> 5a = 2b = 4c và a³ + b³ + c³ = 9512

\(5a=2b=4c\Rightarrow\dfrac{a}{\dfrac{1}{5}}=\dfrac{b}{\dfrac{1}{2}}=\dfrac{c}{\dfrac{1}{4}}\Rightarrow\dfrac{a^3}{\dfrac{1}{125}}=\dfrac{b^3}{\dfrac{1}{8}}=\dfrac{c^3}{\dfrac{1}{64}}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a^3}{\dfrac{1}{125}}=\dfrac{b^3}{\dfrac{1}{8}}=\dfrac{c^3}{\dfrac{1}{64}}=\dfrac{a^3+b^3+c^3}{\dfrac{1}{125}+\dfrac{1}{8}+\dfrac{1}{64}}=\dfrac{9512}{\dfrac{1189}{8000}}=64000\)

\(\Rightarrow\left\{{}\begin{matrix}a^3=\dfrac{64000}{125}=512\\b^3=\dfrac{64000}{8}=8000\\c^3=\dfrac{64000}{64}=1000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=8\\b=20\\c=10\end{matrix}\right.\)

=> A = 8 + 20 + 10 = 38

Ta có:

\(\frac{abc}{a+bc}\)=\(\frac{bca}{b+ca}\)

<=>\(\frac{abc}{bca}=\frac{a+bc}{b+ca}\)

=>\(\frac{a+bc}{b+ca}\)=>\(\frac{a}{b}=\frac{bc}{ca}\)(tính chất dãy tỉ số bằng nhau)<=>\(\frac{a}{bc}=\frac{b}{ca}\)(đpcm)

2009 chia 3 dư 2

10¹⁰ = 10..000 chia 3 dư 1

=> 2009 + 10¹⁰ chia hết cho 3

Vậy 2009 + 10¹⁰ là hợp số

chuối đỏ = chó đuổi => quay về

\(\dfrac{\dfrac{2}{5}}{3x}=\dfrac{\dfrac{2}{3}}{\dfrac{4}{5}}\Leftrightarrow\dfrac{2}{5}\cdot\dfrac{4}{5}=3x\cdot\dfrac{2}{3}\Leftrightarrow\dfrac{8}{25}=3x\cdot\dfrac{2}{3}\Leftrightarrow3x=\dfrac{\dfrac{8}{25}}{\dfrac{2}{3}}=\dfrac{12}{25}\Leftrightarrow x=\dfrac{4}{25}\)

\(\left(x+5\right)+\left(x-9\right)=x+2\)

=> x+5 +x - 9 = x + 2

=> 2x - 4 = x+2

=> 2x - x = 2 + 4

=> x = 6

a.

\(\left(3x-5\right)^2=\dfrac{25}{9}\)

\(\Rightarrow\left[{}\begin{matrix}3x-5=\sqrt{\dfrac{25}{9}}\\3x-5=-\sqrt{\dfrac{25}{9}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-5=\dfrac{5}{3}\\3x-5=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{20}{9}\\x=\dfrac{10}{9}\end{matrix}\right.\)

Vậy:............

b.

\(\left\{{}\begin{matrix}\left(x-2\right)^{2018}\ge0\\\left|y^2-9\right|^{2014}\ge0\\\left(x-2\right)^{2018}+\left|y^2-9\right|^{2014}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2018}=0\\\left|y^2-9\right|^{2014}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=\pm3\end{matrix}\right.\)