\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)

\(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{2^3.3^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}\) =\(2^3.8^5\)

a) Ta có:

\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)

\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)

\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)

Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.

Vậy A > C > B.

b) Ta có:

\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)

\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)

Vậy B : A = -4

a: \(A=\left[6\cdot\dfrac{1}{27}+3\cdot\dfrac{1}{3}+1\right]:\dfrac{-4}{3}\)

\(=\left(\dfrac{2}{9}+2\right)\cdot\dfrac{-3}{4}\)

\(=\dfrac{20}{9}\cdot\dfrac{-3}{4}=\dfrac{-60}{36}=\dfrac{-5}{3}\)

b: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)

\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{5x}{10}=\dfrac{3y}{9}=\dfrac{5x+3y}{10+9}=\dfrac{38}{19}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\end{matrix}\right.\)

b) \(\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x^2}{3^2}=\dfrac{y^2}{5^2}=\dfrac{x^2+y^2}{9+25}=\dfrac{68}{34}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.5=10\end{matrix}\right.\)

c) Nếu phải dùng tính chất của dãy tỉ số bằng nhau thì mình không chắc mình làm đúng, thôi thì:

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

Vì \(x.y=10\) nên \(2k.5k=10\Rightarrow10k^2=10\Rightarrow k^2=1\Rightarrow\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1.2=2\\x=\left(-1\right).2=2\end{matrix}\right.\\\left[{}\begin{matrix}y=1.5=5\\y=\left(-1\right).5=-5\end{matrix}\right.\end{matrix}\right.\)

a: Ta có: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)

\(=35^2-35\cdot13+13^2-35\cdot13\)

\(=\left(35-13\right)^2\)

\(=22^2=484\)

b: Ta có: \(B=\dfrac{68^3-52^3}{16}+68\cdot52\)

\(=68^2+68\cdot52+52^2+68\cdot52\)

\(=\left(68+52\right)^2=14400\)

a,Áp dụng dãy tỉ số bằng nhau ta có

\(\dfrac{x}{5}\)=\(\dfrac{y}{2}\)=\(\dfrac{3x}{15}\)=\(\dfrac{2x}{4}\)=\(\dfrac{3x-2x}{15-4}\)=\(\dfrac{44}{11}\)=4

Suy ra

\(\dfrac{x}{5}\)=4=>x=4x5=20

\(\dfrac{y}{2}\)=4=>y=4x2=8

vậy x=20;y=8

b,Áp dụng dãy tỉ số bằng nhau ta có

2x=3y=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{6}\)=\(\dfrac{x}{3}=\dfrac{y}{2}\)=\(\dfrac{x+y}{3+2}\)=\(\dfrac{10}{5}\)=2

Suy ra:

\(\dfrac{x}{3}\)=2=>x=3x2=6

\(\dfrac{y}{2}\)=2=>y=2x2=4

Vậy x=6,y=4

a,\(\dfrac{x}{2}=\dfrac{y}{3}\) <=> \(\dfrac{5x}{10}=\dfrac{3y}{9}\)

Áp dụng T/c dãy tỉ số BN, ta có:

\(\dfrac{5x+3y}{10+9}=\dfrac{38}{19}=2\). Từ đó suy ra: x=2.10:5=4

y=2.9:3=6

b, \(\dfrac{x}{3}=\dfrac{y}{5}\) <=> \(\dfrac{x^2}{9}=\dfrac{y^2}{25}\)

Áp dụng ......, ta có:

\(\dfrac{x^2+y^2}{9+25}=\dfrac{68}{34}=2\). Từ đó suy ra: x²=2.9=18=>x=..... (xem lại đề)

y²=2.25=50=>y=.... (xem lại đề)

c, \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x.y}{2.5}=\dfrac{10}{10}=1\)

=> x=1.2=2

y=1.5=5