Áp dụng |A| + |B| \(\ge\)|A + B|

|x - 2| + |x - 3| = |x - 2| + |3 - x|\(\ge\)|x - 2 + 3 - x| = 1

Vậy GTNN của x = 1 khi và chỉ khỉ x \(\ge\)3

\(B=\left|x-3\right|+\left|x-2\right|\)

\(\Rightarrow B=\left|x-3\right|+\left|2-x\right|\)

\(\ge\left|\left(x-3\right)+\left(2-x\right)\right|\)

\(=\left|-1\right|=1\)

Vậy \(B_{min}=1\Leftrightarrow\left(x-3\right)\left(2-x\right)\ge0\)

\(TH1:\hept{\begin{cases}x-3\ge0\\2-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le2\end{cases}}\left(L\right)\)

\(TH2:\hept{\begin{cases}x-3\le0\\2-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge2\end{cases}}\Leftrightarrow2\le x\le3\)

Vậy dấu "="\(\Leftrightarrow\)\(2\le x\le3\)

a) Vì AD là phân giác BAC 

=> BAD = CAD = \(\frac{60°}{2}\)= 30° 

Vì AB//DM 

=> BAD = ADM = 30° ( so le trong) 

Vì AD//MK 

=> ADM = DMK = 30° ( so le trong) 

b) Mà AD//MK 

=> DAC = KMC = 30° ( đồng vị) 

=> DMK = KMC = 30° 

Hay MK là phân giác DMC 

mn ơi giúp mình với huhu

3.5.7.11.12.37-10101/1212120+40404

=512820-10101/1252524

=502719/1252524

=647/1612

mình ko chắc chắn lăm

\(ac-a^2+bc-ab=ac+a^2-bc-ab\)​​

\(\Leftrightarrow-a^2+bc-ab=a^2-bc-ab\)

\(\Leftrightarrow-a^2+bc=a^2-bc\)

\(\Leftrightarrow2a^2=2bc\)

\(\Leftrightarrow a^2=bc\)

a, \(\frac{2^{15}.\left(-9\right)^4}{-6^3.8^3}=\frac{2^{15}.\left(-3.3\right)^4}{-\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^4.3^4}{-2^3.3^3.2^9}=\frac{2^{15}.3^8}{-2^{12}.3^3}=\frac{2^3.3^5}{-1}=-8.243=-1944\)

b, \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=\frac{1}{2^8}=\frac{1}{256}\)

\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

\(\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)

\(\Leftrightarrow ac+bc-ad-bd=ac-bc+ad-bd\)

\(\Leftrightarrow ac+bc-ad-bd-ac+bc-ad+bd=0\)

\(\Leftrightarrow2bc-2ad=0\)

\(\Leftrightarrow2bc=2ad\)

\(\Leftrightarrow bc=ad\)

\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)(đpcm)