Ta có :

a < b \(\Rightarrow\)2a < a + b \(\Rightarrow\)\(\frac{a}{a+b}< \frac{1}{2}\)

c < d \(\Rightarrow\)2c < c + d \(\Rightarrow\)\(\frac{c}{c+d}< \frac{1}{2}\)

m < n \(\Rightarrow\)2m < m + n \(\Rightarrow\)\(\frac{m}{m+n}< \frac{1}{2}\)

\(\Rightarrow\)2a + 2c + 2m < ( a + b ) + ( c + d ) + ( m + n ) 

\(\Rightarrow\)2 . (a  + c + nm ) < a + b + c + d + m + n

\(\Rightarrow\)\(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

\(a< b\Rightarrow2a< a+b\)

\(c< d\Rightarrow2c< c+d\)

\(m< n\Rightarrow2m< m+n\)

\(\Rightarrow2a+2c+2m< a+b+c+d+m+n\)

\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(\text{đ}pcm\right)\)

a) vì \(\left|x+\frac{15}{19}\right|\ge0\text{ }\forall\text{ }x\)

\(\Rightarrow\)M_min  \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = \(\frac{-15}{19}\)

b) vì \(\left|x-\frac{4}{7}\right|\ge0\text{ }\forall\text{ }x\)

\(\Rightarrow\)\(\left|x-\frac{4}{7}\right|-\frac{1}{2}\ge\frac{-1}{2}\)

\(\Rightarrow\)N_min \(\Leftrightarrow\)N = \(\frac{-1}{2}\)\(\Rightarrow\)\(x=\frac{4}{7}\)

không cần SKT_NTT trả lời

a) vì | x + 15/19 | \(\ge\)0 \(\forall\)x

\(\Rightarrow\)M_min \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = -15/19

b) vì | x - 4/7 | \(\ge\)0 \(\forall\)x

\(\Rightarrow\)|x  - 4/7 | - 1/2 \(\ge\)-1/2

\(\Rightarrow\)N_min \(\Leftrightarrow\)N = -1/2 \(\Rightarrow\)x = 4/7

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)

\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)

\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)

\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)

Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)

ko biet

chịu ko bít

uoooo

Đéo biết thì đừng có nói

cách giải là

\(\frac{4}{9}\)và \(\frac{13}{18}\)\(\Rightarrow\frac{4}{9}=\frac{4.2}{9.2}=\frac{8}{18}\)\(,\frac{13}{18}\)GIỮ NGUYÊN 

VÌ \(\frac{8}{18}< \frac{13}{18}\)NÊN \(\frac{4}{9}< \frac{13}{18}\)

\(\frac{-15}{7}\)VÀ \(\frac{-6}{5}\)\(\Rightarrow\frac{-15}{7}=\frac{-15.5}{7.5}=\frac{-75}{35}\)

                                 \(\frac{-6}{5}=\frac{-6.7}{5.7}=\frac{-42}{35}\)

VÌ \(\frac{-75}{35}< \frac{-42}{35}\)    NÊN    \(\frac{-15}{7}< \frac{-6}{5}\)

MK CHẮC CHẮN SẼ ĐÚNG

\(\frac{4}{9}< \frac{13}{18}\)

\(\frac{-15}{7}< \frac{-6}{5}\)