A

\(2400:500=4\) dư \(2400-500.4=400\Rightarrow C\)

\(\dfrac{\left(\dfrac{2}{7}\right)^7.7^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\)

\(=\dfrac{\dfrac{2^7}{7^7}.7^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7.5^2+2.256}\)

\(=\dfrac{2^7+\left(\dfrac{9}{4}.\dfrac{16}{3}\right)^3}{2^7.5^2+2.2^8}=\dfrac{2^7+\left(12\right)^3}{2^7.5^2+2.2^8}\)

\(=\dfrac{2^7+\left(2^2.3\right)^3}{2^7.5^2+2^9}=\dfrac{2^7+2^6.3^3}{2^7.\left(5^2+2^2\right)}\)

\(=\dfrac{2^6\left(2+27\right)}{2^7.\left(25+4\right)}=\dfrac{29}{2.29}=\dfrac{1}{2}\)

\(\dfrac{3x-1}{3x^2+5x+2}=\dfrac{1}{x+2}\left(x\ne-2;x\ne\dfrac{1}{3}\right)\)

\(\Rightarrow\left(3x-1\right)\left(x+2\right)=3x^2+5x+2\)

\(\Rightarrow3x^2+6x-x-2=3x^2+5x+2\)

\(\Rightarrow3x^2+5x-2=3x^2+5x+2\)

\(\Rightarrow-2=2\) (vô lý)

Bạn xem lại đề bài

\(5^{36}=\left(5^3\right)^{12}=125^{12}\)

\(11^{24}=\left(11^2\right)^{12}=121^{12}< 125^{12}\)

\(\Rightarrow5^{36}>11^{24}\)

\(^{^{ }}\)5³⁶>11²⁴nhé

a) Thể tích hình hộp bị khuyết :

\(4.4.4=4^3=64\left(cm^3\right)\)

b) Diện tích xung quanh hình hộp bị khuyết (3 mặt)

\(3.4.4.4=192\left(cm^2\right)\)

làm như bạn trí là đúng nhé

\(2x^2+8xy+6y^2=24\left(x;y\inℕ^∗\right)\)

\(\Rightarrow2\left(x^2+4xy+3y^2\right)=24\)

\(\Rightarrow\left(x^2+4xy+4y^2\right)-y^2=12\)

\(\Rightarrow\left(x+2y\right)^2-y^2=12\)

\(\Rightarrow\left(x+2y+y\right)\left(x+2y-y\right)=12\)

\(\Rightarrow\left(x+3y\right)\left(x+y\right)=12\)

\(\Rightarrow\left(x+3y\right);\left(x+y\right)\in\left\{1;2;3;4;6;12\right\}\)

\(\Rightarrow\left(x+3y\right);\left(x+y\right)\in\left\{1;2;3;4;6;12\right\}\)

Ta giải các hệ phương trình sau :

1) \(\left\{{}\begin{matrix}x+3y=1\\x+y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=-11\left(loại\right)\\x+y=12\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}x+3y=2\\x+y=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=-3\left(loại\right)\\x+y=6\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}x+3y=3\\x+y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=-1\left(loại\right)\\x+y=4\end{matrix}\right.\)

4) \(\left\{{}\begin{matrix}x+3y=4\\x+y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=1\left(loại\right)\\x+y=3\end{matrix}\right.\)

5) \(\left\{{}\begin{matrix}x+3y=6\\x+y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=4\\x+y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\x=0\left(loại\right)\end{matrix}\right.\)

6) \(\left\{{}\begin{matrix}x+3y=12\\x+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=11\left(loại\right)\\x+y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\varnothing\right\}\)