\(\dfrac{25}{100}\times30+2,5:10+\dfrac{1}{4}\times29+0,25\times40\)

\(=0,25\times30+0,25\times1+0,25\times29+0,25\times40\)

\(=0,25\times\left(30+1+29+40\right)\)

\(=0,25\times100\)

\(=25\)

\(3,27\times0,25+2,3\times\dfrac{1}{4}+1,75:4+2,68\times25\%\)

\(=3,27\times0,25+2,3\times0,25+1,75\times0,25+2,68\times0,25\)

\(=\left(3,27+2,3+1,75+2,68\right)\times0,25\)

\(=\left(5,57+1,75+2,68\right)\times0,25\)

\(=\left(7,32+2,68\right)\times0,25\)

\(=10\times0,25\)

\(=2,5\)

\(\dfrac{154,8+42,7}{x}+35=74,5\)

\(\Rightarrow\dfrac{197,5}{x}+35=74,5\)

\(\Rightarrow\)\(\dfrac{197,5}{x}=74,5-35\)

\(\Rightarrow\dfrac{197,5}{x}=39,5\)

\(\Rightarrow x=197,5:39,5\)

\(\Rightarrow x=5\)

\(A=1+2^1+2^2+2^3+2^4+2^5\)

\(1,\)

\(2A=2+2^2+2^3+2^4+2^5+2^6\)

\(2A-A=\left(2+2^2+2^3+2^4+2^5+2^6\right)-\left(1+2^1+2^2+2^3+2^4+2^5\right)\)

\(A=2^6-1\)

\(A=64-1=63\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

Ta có :

\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

\(\Rightarrow xy=54\Leftrightarrow2k.3k=54\)

\(\Rightarrow6k^2=54\)

\(\Rightarrow k^2=54:6\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k^2=\left(\pm3\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(\pm3\right)=\left(\pm6\right)\\y=3.\left(\pm3\right)=\left(\pm9\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(6;9\right),\left(-6;-9\right)\right\}\)

 Mình có thấy cái đề bài đâu nhỉ????????

\(8,64\times46+52\times8,64+17,28\)

\(=8,64\times46+52\times8,64+8,64\times2\)

\(=8,64\times\left(46+52+2\right)\)

\(=8,64\times100=864\)

\(26+5x=3x+56\)

\(\Rightarrow5x-3x=56-26\)

\(\Rightarrow2x=30\)

\(\Rightarrow x=30:2\)

\(\Rightarrow x=15\)

Vậy \(x=15\)

 Tớ không thấy đề bài cậu ơi!

\(24.\left[18-\left(3^2-2\right)^2\right]\)

\(=24.\left[18-\left(9-2\right)^2\right]\)

\(=24.\left[18-7^2\right]\)

\(=24.\left[18-49\right]\)

\(=24.-31=-744\)