pt \(\lrArr\) \(2^{2x}\) - 12x\(2^{x}\) + m = 0

đặt \(2^{x}\) = t \(\rarr\) x = \(\log_2^{}t\)

pt \(\lrArr\) \(t^2\) + 12t + m =0

để pt có 2 nghiệm pb \(\rarr\) \(\Delta\) \(\) > 0

\(\lrArr\) 144 - 4m > 0

\(\lrArr\) m < 36

có \(x_1^{}\) + \(x_2^{}\) = 5

\(\lrArr\) \(\log_2^{}t_1^{}\) + \(\log_2^{}t_2^{}\) = 5

\(\lrArr\) \(\log_2^{}t_1^{}t_2^{}\) = 5

\(\lrArr\) \(t_1^{}t_2^{}\) = 32

\(\lrArr\) m = 32 (tm)

a) p(\(\overline{A}B\) ) = 0.8 x 0.3 = 0.24

b) p(\(\overline{A}\cup\overline{B}\) ) = 0.8 + 0.7 - 0.8x0.7 = 0.94

\(\frac{d\left(D,\left(SBM\right)\right)}{d\left(A,\left(SBM\right)\right)}\) = \(\frac12\)

kẻ AK \(\) \(\)⊥ BM, AH ⊥ SK

có BM ⊥ AK, BM ⊥ SA \(\rarr\) BM ⊥ (SAK) \(\rarr\) BM ⊥ AH

có AH ⊥ SK, BM ⊥ AH \(\rarr\) AH ⊥ (SBM)

\(\rarr\) d(A,(SBM)) = AH

kẻ AD \(\cap\) BM tại E

mà AB // DM \(\rarr\) theo talet ta có: \(\frac{AB}{DM}\) = \(\frac{AE}{DE}\) \(\lrArr\) AE = 2a

có \(\Delta\)ABE vuông tại A \(\rarr\) \(\frac{1}{AK^2}\) = \(\frac{1}{AB^2}\) + \(\frac{1}{AE^2}\) \(\lrArr\) AK = \(\frac{2a\sqrt5}{5}\)

có \(\Delta\)SAK vuông tại A \(\rarr\) \(\frac{1}{AH^2}\) = \(\frac{1}{AK^2}\) + \(\frac{1}{SA^2}\) \(\lrArr\) AH = \(\frac{a\sqrt6}{3}\)

\(\rarr\) d(D,(SBM)) = \(\frac{AH}{2}\) = \(\frac{a\sqrt6}{6}\)