\(\frac{4^{10}+8^4}{4^5+8^5}=\frac{\left(2^2\right)^{10}+\left(2^3\right)^4}{\left(2^2\right)^5+\left(2^3\right)^5}=\frac{2^{20}+2^{12}}{2^{10}+2^{15}}\)

                    \(=\frac{2^{12}\left(2^8+1\right)}{2^{10}\left(1+2^5\right)}=\frac{2^{12}.257}{2^{10}.33}=\frac{2^2.257}{33}=\frac{1028}{33}\)    

Ta có: \(\frac{x}{2}=\frac{y}{3}\)

\(=>\frac{6x}{12}=\frac{4y}{12}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có: \(\frac{6x+4y}{12+12}=\frac{72}{24}=3\)

\(=>\hept{\begin{cases}\frac{6x}{12}=3\\\frac{4y}{12}=3\end{cases}=>\hept{\begin{cases}x=6\\y=9\end{cases}}}\)

Vậy...

Ta có: \(\frac{x}{3}=\frac{y}{5}\)

\(=>\frac{x^2}{9}=\frac{y^2}{25}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có: \(\frac{x^2-y^2}{9-25}=\frac{-4}{-16}=\frac{1}{4}\)

\(=>\hept{\begin{cases}\frac{x^2}{9}=\frac{1}{4}\\\frac{y^2}{25}=\frac{1}{4}\end{cases}=>\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{5}{2}\end{cases}}}\)

Vậy:...

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

\(A=\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy .......

Haiz, sao lại thiếu sự quan sát thế nhỉ?

TH1: \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)\(\Rightarrow A=\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}=-1\)

TH2: \(a+b+c\ne0\)\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(2^x+2^{x+4}=544\)

\(\Leftrightarrow2^x+2^x.2^4=544\)

\(\Leftrightarrow2^x.17=544\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow x=5\)

Ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

=> \(\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\left(\frac{a}{b}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (Vì \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\))

=> \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\)(đpcm)

Ta có : (x + 1) : (14 - x) = 2:3

=> \(\frac{x+1}{14-x}=\frac{2}{3}\)

=> 3(x + 1) = 2(14 - x)

=> 3x  + 3 = 28 - 2x

=> 5x = 25

=> x = 5

Vậy x = 5 là giá trị cần tìm