Ta có : 81^x = 27 

\(\Leftrightarrow\)( 3⁴ )^x = 27

\(\Rightarrow\)3^4x = 3³

\(\Rightarrow\)4x = 3

\(\Rightarrow\)x = 3 / 4

Vậy : x = 3 / 4

81^x=27

(3⁴)^x=3³

3^4x=3³

=>4x=3

      x=3/4

vậy x=3/4

a) Ta có: 3|x - 14| \(\ge\)0 \(\forall\)x

=> 3|x - 14| + 4 \(\ge\)4 \(\forall\)x

=> \(\frac{6}{3\left|x-14\right|+4}\le\frac{3}{2}\forall x\)

Dấu "=" xảy ra <=> x - 14 = 0 <=> x = 14

Vậy MaxA = 3/2 <=> x = 14

b) Mình có: |2x + 6| = \(\orbr{\begin{cases}2x+6\\-2x-6\end{cases}}\)\(\Rightarrow\)B_Min = - 2x- 6  + 2 + 2x = -4 khi x \(\le\)-3

Ta có: \(B=\frac{1}{112}-\frac{1}{84}-\frac{1}{60}-\frac{1}{40}-\frac{1}{24}-\frac{1}{12}-\frac{1}{4}\)

\(\Rightarrow2B=\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\frac{6}{7}\)

\(\Rightarrow2B=-\frac{47}{56}\)

\(\Rightarrow B=-\frac{47}{112}\)

Hok tốt nha^^