\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7:\left(\frac{3}{5}\right)^2\)

\(x=\frac{3^7}{7^7}:\frac{3^2}{5^2}\)

\(x=\frac{3^7.5^2}{7^7.3^2}\)

\(x=\frac{3^5.5^2}{7^7}=\frac{6075}{823543}\)

c) Theo bài ra: \(\frac{2x}{3}=\frac{y}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{5y}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{\frac{3}{2}}=\frac{5y}{25}=\frac{x+5y}{\frac{3}{2}+25}=\frac{3}{\frac{53}{2}}=\frac{6}{53}\)

\(\Rightarrow x=\frac{9}{53};y=\frac{30}{53}\)

a) Ta có: \(\frac{x}{-5}=\frac{2y}{3}\Rightarrow\frac{x}{-5}=\frac{y}{\frac{3}{2}}\Rightarrow\frac{x}{-5}=\)\(\frac{5y}{\frac{15}{2}}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{-5}=\frac{5y}{\frac{15}{2}}=\frac{x+5y}{\left(-5\right)+\frac{15}{2}}=\)\(\frac{-1}{\frac{5}{2}}=\frac{-2}{5}\)

\(\Rightarrow x=2;y=-0,6\)

a)    (a+b)-(-c+a+b)

=      a+b+c-a-b

= c

b)     -(x+y)+(-z+x+y)

=       -x-y-z+x+y

=       x

c)      ( m-n+p)+(-m+n+p)

=       m-n+p-m+n+p

=       p+p

=      2p

a,a+b+c-a-b

= c

b,= -x-y-z+x+y

= -z

c,= m-n+p-m+n+p

= 2p

\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4}}}}\)

\(=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{5}{4}}}}\)

\(=1+\frac{1}{1+\frac{1}{1+\frac{4}{5}}}\)

\(=1+\frac{1}{1+\frac{1}{\frac{9}{5}}}\)

\(=1+\frac{1}{1+\frac{5}{9}}\)

\(=1+\frac{1}{\frac{14}{9}}\)

\(=1+\frac{9}{14}\)

\(=\frac{23}{14}\)

\(\frac{a}{5}=\frac{b}{4}\Rightarrow a=5k\) ; \(b=4k\)

Ta có : \(5k+4k=36\Rightarrow9k=36\Rightarrow k=4\)

\(\Leftrightarrow\hept{\begin{cases}\frac{a}{5}=4\\\frac{b}{4}=4\end{cases}\Rightarrow}\hept{\begin{cases}a=20\\b=16\end{cases}}\)

Vậy a = 20 ; b = 16 

a/5 + b/4 => a = 5k ; b = 4k

Ta co : 5k + 4k = 36 => 9k = 36 => k = 6

a/5 = 4 => a = 20

b/4 = 4 => b = 16

ta có BE=BC nên ​\(\Delta\)BEC cân => \(\widehat{E}\)=\(\widehat{BCE}\)

\(\widehat{ABC}\)là góc ngoài của tg BEC nên \(\widehat{ABC}\)=\(\widehat{E}\)+\(\widehat{BCE}\)= \(\widehat{2E}\)

Mà \(\widehat{ABD}=\widehat{DBC}\)=>  \(\widehat{E}\)=\(\widehat{BCE}\)=\(\widehat{ABD}=\widehat{DBC}\)=> DB//CE

Nếu bn hok đồng dư r thì dễ lắm, dư 0 phải hok

Sorry là BD=BC nhé!

Chịu thôi hình kiểu này sao tính được CD ?