a) xⁿ.x^m = x^{n + m}

b) \(\frac{x^n}{x^m}=x^n:x^m=x^{n-m}\)

c) \(\left(x^n\right)^m=x^{n.m}\)

d) \(\left(x.y\right)^n=x^n,y^n\)

e) \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\)

\(\frac{3}{4}\left(\frac{2}{5}\right)^{14}:\left(\frac{4}{25}\right)^6=\frac{3}{4}\left(\frac{2}{5}\right)^{14}:\left(\frac{2}{5}\right)^{2.6}=\frac{3}{4}\left(\frac{2}{5}\right)^2=\frac{3}{4}.\frac{4}{25}=\frac{3}{25}\)

a) \(x^n.x^m=x^{n+m}\)

b) \(\frac{x^n}{x^m}=x^n\div x^m=x^{n-m}\)

c) \(\left(x^n\right)^m=x^{n.m}\)

d) \(\left(x.y\right)^n=x^n.y^n\)

e) \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\left(y\ne0\right)\)

Áp dụng \(\frac{3}{4}.\left(\frac{2}{5}\right)^{14}\div\left(\frac{4}{25}\right)^6=\frac{3}{4}.\frac{2^{14}}{5^{14}}\div\frac{4^6}{25^6}\)

                                                          \(=\frac{3}{4}.\frac{2^{14}}{5^{14}}.\frac{25^6}{4^6}\)

                                                          \(=\frac{3.2^{14}.\left(5^2\right)^6}{4.5^{14}.\left(2^2\right)^6}=\frac{3.2^{14}.5^{12}}{2^2.5^{14}.2^{12}}=\frac{3}{25}\)

a) Xét tứ giác BECD có : 

M là trung điểm ED 

M là trung điểm BC

=》 BECD là hình bình hành 

=》BE//DC 

b) Vì BECD là hình bình hành 

=》EC//BD 

Mà NBD = 90°

Lại có : NBD + CNB = 180°

=》 CNB = 90°

Vậy CN\(\perp\)AB 

Hay CE\(\perp\)AB

Bạn làm lại theo cách làm của hình Tam Giác giúp mình

a) Ta có: \(A=\left|3x+\frac{1}{3}\right|-\frac{1}{4}\ge-\frac{1}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|3x+\frac{1}{3}\right|=0\Rightarrow x=-\frac{1}{9}\)

Vậy Min(A) = -1/4 khi x = -1/4

b) Ta có: \(\frac{3}{4}-\left|2x-\frac{1}{2}\right|0\le\frac{3}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|2x-\frac{1}{2}\right|=0\Rightarrow x=\frac{1}{4}\)

Vậy Max(B) = 3/4 khi x = 1/4

a. Vì \(\left|3x+\frac{1}{3}\right|\ge0\forall x\)\(\Rightarrow A=\left|3x+\frac{1}{3}\right|-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|3x+\frac{1}{3}\right|=0\Leftrightarrow3x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{9}\)

Vậy minA = - 1/4 <=> x = - 1/9

b. Vì \(\left|2x-\frac{1}{2}\right|\ge0\forall x\)\(\Rightarrow B=\frac{3}{4}-\left|2x-\frac{1}{2}\right|\le\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|2x-\frac{1}{2}\right|=0\Leftrightarrow2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy maxB = 3/4 <=> x = 1/4

Bài làm:

Ta có: \(a=15^{120}\div25^{60}\)

\(a=15^{120}\div5^{120}\)

\(a=3^{120}=9^{60}\)

và \(b=2^{45}.2^{15}.4^{60}\)

\(b=2^{60}.2^{120}\)

\(b=2^{180}=8^{60}\)

Mà \(9^{60}>8^{60}\Rightarrow a>b\)

    \(\frac{1}{2.15}+\frac{3}{2.11}+\frac{4}{1.11}+\frac{5}{1.2}\)

\(=\frac{1}{30}+\left(\frac{3}{22}+\frac{4}{11}\right)+\frac{5}{2}\)

\(=\frac{1}{30}+\frac{1}{2}+\frac{5}{2}\)

\(=\frac{1}{30}+3\)

\(=\frac{91}{30}\)

= 3 \(\frac{1}{30}\)Hoặc  =\(\frac{91}{30}\)

\(\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab\)    

\(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)     

VT = \(\frac{\left(b-a\right)\left(b+a\right)}{a^2+ab}\)      

=\(\frac{\left(b-a\right)\left(b+a\right)}{a\left(a+b\right)}\)      

=\(\frac{b-a}{a}\)        

= VP 

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

=> \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)

=> \(1-\frac{1}{2n+1}=\frac{50}{51}\)

=> \(\frac{1}{2n+1}=1-\frac{50}{51}=\frac{1}{51}\)

=> 2n + 1 = 51 

=> 2n = 50

=> n = 25

Vậy n = 25