\(\frac{\left(3.4.2^4\right)^2}{5.2^5.4^2-16^2}\)

\(=\frac{\left(3.2^2.2^4\right)^2}{5.2^5\left(2^2\right)^2-\left(2^4\right)^2}\)

\(=\frac{\left(3.2^6\right)^2}{5.2^9-2^8}\)

\(=\frac{3^2.2^{12}}{2^8\left(5.2-1\right)}\)

\(=\frac{9.2^4}{9}\)

\(=16\)

\(=\frac{3^2.2^4.2^8}{5.2^5.2^4-2^8}\)

\(=\frac{3^22^{12}}{2^8\left(5.2-1\right)}\)\(=2^4\)

9/160

chép nhầm đề. Làm lại:

\(C=\frac{\left(3.4.2^4\right)^2}{5.2^5.4^2-16^2}=\frac{3^2.2^4.2^8}{5.2^5.2^4-2^8}=\frac{9.2^{12}}{5.2^9-2^8}=\frac{9.2^{12}}{2^8.\left(5.2-1\right)}=\frac{9.2^4}{9}=\frac{2^4}{1}=2^4=16\)

Gấp lắm thì làm vậy

\(C=\frac{\left(3.4.2^2\right)^2}{5.2^5.4^2-16^2}=\frac{3^2.2^4.2^4}{5.2^5.2^4-2^8}=\frac{3^2.2^8}{5.2^9-2^8}=\frac{3^2.2^8}{2^8.\left(5.2-1\right)}=\frac{9}{9}=1\)

\(B=\dfrac{x+16-5\sqrt{x}-10}{x-4}=\dfrac{x-5\sqrt{x}+6}{x-4}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

Bài 1:

a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)

\(=\frac{-15}{240}-\frac{16}{240}\)

\(=\frac{-31}{240}\)

b, \(=\frac{-10}{12}-\frac{-12}{12}\)

\(=\frac{2}{12}=\frac{1}{6}\)

c, \(=\frac{-30}{6}-\frac{1}{6}\)

\(=\frac{-31}{6}\)

Bài 2:

a, \(x=-\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

b,   \(\frac{1}{2}+x=-\frac{11}{2}\)

\(x=-\frac{11}{2}-\frac{1}{2}\)

\(x=-6\)

Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^