\(A=\dfrac{\left(-3\right)^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)^{45}\cdot\left(-2\right)^{12}}{5\cdot\left(-3\right)^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)\cdot\left(-2\right)}{5}=\dfrac{6}{5}\)

Thank u very much!!

Bài 1:
\(130050:452=287\)(dư 326)
\(19183:78=245\)(dư 73)
\(204\times1942=396168\)
Bài 2:
\(\dfrac{4}{9}+\dfrac{3}{7}=\dfrac{28}{63}+\dfrac{27}{63}=\dfrac{55}{63}\)
\(\dfrac{7}{15}-\dfrac{11}{30}=\dfrac{14}{30}-\dfrac{11}{30}=\dfrac{3}{30}=\dfrac{1}{10}\)

a) \(k=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.\left(3^2\right)^2}{3^5.\left(2^4\right)^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{8.1}{3.1}=\frac{8}{3}\)

b)  \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{\left(3^2\right)^3.\left(3^3\right)^2}{\left(2.3\right)^2.3^{10}}=\frac{3^6.3^6}{2^2.3^2.3^{10}}=\frac{3^{12}}{4.3^{12}}=\frac{1}{4}\)

a) \(K=\frac{2^{11}\cdot9^2}{3^5\cdot16^2}=\frac{2^{11}\cdot3^4}{3^5\cdot2^8}=\frac{2^3}{3}=\frac{8}{3}\)

b) \(N=\frac{9^3\cdot27^2}{6^2\cdot3^{10}}=\frac{3^6\cdot3^6}{2^2\cdot3^2\cdot3^{10}}=\frac{1}{4}\)

c) \(P=\frac{27^{15}\cdot5^3\cdot8^4}{25^2\cdot81^{11}\cdot2^{11}}=\frac{3^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot2^{11}}=\frac{3\cdot2}{5}=\frac{6}{5}\)

dấu . là nhân hay là phần ngăn cách ở hàng phần nghìn thế

DẤU CHẤM LÀ PHẦN NGĂN CÁCH 

a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

mình cảm ơn

1) \(\dfrac{x}{3}=\dfrac{y}{4}=t\Leftrightarrow\left\{{}\begin{matrix}x=3t\\y=4t\end{matrix}\right.\)

ta có \(x.y^2=324\Leftrightarrow3t.\left(4t\right)^2=324\)

\(\Leftrightarrow t^3=\dfrac{27}{4}\)

\(\Leftrightarrow t=\dfrac{3}{\sqrt[3]{4}}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3.\dfrac{3}{\sqrt[3]{4}}=\dfrac{9}{\sqrt[3]{4}}\\y=4.\dfrac{3}{\sqrt[3]{4}}=\dfrac{12}{\sqrt[3]{4}}\end{matrix}\right.\)

2) \(2^{x+1}.3^y=2^{2x}.3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

3) \(\dfrac{a}{b}=\dfrac{c}{d}\)

áp dụng dãy tỉ số = nhau ta có

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\)

\(\Leftrightarrow\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\left(\dfrac{a-c}{b-d}\right)^4\left(1\right)\)

mà \(\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\dfrac{a^4+c^4}{b^4+c^4}\left(2\right)\)

từ (1)(2) suy ra đpcm

4) \(B=\dfrac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{3.2}{5}=\dfrac{6}{5}\)

\(\frac{y}{12}=\frac{x}{4}=\frac{y-x}{12-4}=\frac{4}{8}=\frac{1}{2}.\)

Từ đó tính được x và y => Z

Áp dụng tính chất của dãy tỉ số bằng nhau ta được : 

\(\frac{x}{4}=\frac{y}{12}=\frac{y-x}{12-4}=\frac{4}{8}=\frac{1}{2}\)

Do đó : \(\hept{\begin{cases}\frac{x}{4}=\frac{1}{2}\\\frac{y}{12}=\frac{1}{2}\\\frac{z}{15}=\frac{1}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=6\\z=7,5\end{cases}}\)

Vậy .........

\(1+2+3+4+5+...+28+29=\left(1+29\right).29:2=435\)

(1+29)+(2+28)+(3+27)+(4+26)+(5+25)+(6+24)+(7+23)+(8+22)+(9+21)+(10+20)+(11+19)+(12+18)+(13+17)+(14+16)+15

=30+30+30+30+30+30+30+30+30+30+30+30+30+30+15

=30x14+15

=435