\(\frac34-\left|2x+1\right|=\frac78\)

\(\left|2x+1\right|=\frac34-\frac78\)

\(\left|2x+1\right|=-\frac18\) (vô lý vì \(\left|2x+1\right|\ge0\forall x\) )

⇒ x thuộc rỗng

      \(4^{15}.9^{15}< 2^n.3^n< 18^{16}.2^{16}\)

\(\Rightarrow\left(4.9\right)^{15}< \left(2.3\right)^n< \left(18.2\right)^{16}\)

\(\Rightarrow36^{15}< 6^n< 36^{16}\)

\(\Rightarrow\left(6^2\right)^{15}< 6^n< \left(6^2\right)^{16}\)

\(\Rightarrow6^{30}< 6^n< 6^{32}\Rightarrow30< n< 32\)

Mà n là số tự nhiên nên n = 31

Chúc bạn học tốt.

a. \(4^{15}.9^{15}< 2^n.3^n< 18^{16}.2^{16}\)

\(\Rightarrow2^{30}.3^{30}< 2^n.3^n< \left(3^2\right)^{16}.2^{16}.2^{16}\)

\(\Rightarrow2^{30}.3^{30}< 2^n.3^n< 3^{32}.2^{32}\)

\(\Rightarrow30< n< 32\)

\(\Rightarrow n=31\)

Vậy : \(n=31\)

\(n=0\Rightarrow b=3\)

Với \(n\ne0\Rightarrow VP⋮2butVT\) ko chia hết cho 2 nên ko thỏa mãn

Vậy \(n=0;b=3\)

C.\(\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^6}{2^{5+}2^5}=\frac{4^6}{3^6}.\frac{6^6}{2^5+2^5}=\frac{24^6}{3^6.\left(2^5+2^5\right)}=\frac{8^6}{2^5.\left(1+1\right)}\)=\(\frac{8^6}{2^6}\)=4^6=4096

\(^{4^6=2^{12}}\)

           \(\Rightarrow\)n=12

b) = \(\frac{3}{4}\div\)\(\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)

= \(\frac{3}{4}\div\frac{5}{6}\)

= \(\frac{9}{10}\)

c) \(\frac{16.2^3}{4}\)

\(=4.8=32\)

\(a)\left|-\frac{1}{2}\right|+3^0+\frac{1}{4}+4+2021^0.\)

\(=\frac{1}{2}+1+\frac{1}{4}+4+1\)

\(=\left(\frac{1}{2}+\frac{1}{4}\right)+\left(1+4+1\right)\)

\(=\frac{3}{4}+6=\frac{27}{4}\)

\(b)\frac{3}{4}\div\left(-\frac{1}{3}\right)+\frac{3}{4}\div\frac{2}{3}+\frac{3}{4}\div\frac{1}{2}\)

\(=\frac{3}{4}\div\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)

\(=\frac{3}{4}\div\frac{5}{6}=\frac{9}{10}\)

\(\left(4.9\right)^{15}<\left(2.3\right)^n<\left(18.2\right)^{16}\)

\(36^{15}<6^n<36^{16}\)

\(\left(6^2\right)^{15}<6^n<\left(6^2\right)^{16}\)

\(6^{30}<6^n<6^{32}\)

\(\Rightarrow30

\(\Rightarrow n=31\)