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^^

+Vì Om vuông góc với Ox

=> mOx là góc vuông

=> mOx = 90^o

+Vì On vuông góc với Oy

=> nOy là góc vuông

=> nOy = 90^o

+Ta có : mOx < xOy (90^o<120^o)

=>Om nằm giữa Oy và Ox 

=> mOx + mOy = xOy

=> mOy = 30^o

+ Ta có : mOy < nOy (30^o < 90^o)

=> Om nằm giữa Oy và On

=> mOy + mOn = nOy

=> mOn = 60^o

n mu2  .n .nmu5 =7mu11:7mu3