a) \(|2x-1|\ge0\forall x\)

\(\Rightarrow A=5-|2x-1|\le5\forall x\)

\(A=5\Leftrightarrow|2x-1|=0\)\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Vậy Max A = 5 <=> x = 1/2

b) \(|x-2|\ge0\forall x\)\(\Rightarrow|x-2|+3\ge3\forall x\)

\(\Rightarrow B=\frac{1}{|x-2|+3}\le\frac{1}{3}\forall x\)

\(B=\frac{1}{3}\Leftrightarrow|x-2|=0\)\(\Leftrightarrow x=2\)

Vậy Max B = 1/3 <=>  x = 2 

a) \(A=5-\left|2x-1\right|\)

Ta có  \(2x-1\le0\)

\(\Rightarrow5-\left|2x-1\right|\le5\)

Để  A đạt GTLN   \(\Leftrightarrow x=\frac{1}{2}\)

b) \(B=\frac{1}{\left|x-2\right|+3}\)

Ta có  : \(\left|x-2\right|+3\ge3\)

và \(1>0\)

\(\Rightarrow\frac{1}{\left|x-2\right|+3}\le\frac{1}{3}\)

Để B đạt GTLN \(\Leftrightarrow x=2\)

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

\(bd=1\Rightarrow b=\frac{1}{d}\)

\(\Rightarrow\frac{a}{b}=\frac{1}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{a}{b}=\frac{1}{d}=\frac{a+1}{b+d}\left(đpcm\right)\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)

\(\Rightarrow x+1=0\)( do \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\))

\(\Rightarrow x=-1\)

a) \(3\cdot x^9=x^7\cdot75\)\(\Rightarrow x^9:x^7=75:3\)

\(\Rightarrow x^2=25\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

b) \(\frac{x}{5}=\frac{y}{3}\)\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)( Theo tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\hept{\begin{cases}x^2=\frac{1}{4}\cdot25=\frac{25}{4}\\y^2=\frac{1}{4}\cdot9=\frac{9}{4}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{3}{2}\end{cases}}\)( do x,y > 0 )

\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}=\left(\frac{1}{2}\right)^3\)

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

Vậy,........

\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)

\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)

\(\Rightarrow2n-1=3\)

\(\Rightarrow2n=4\Rightarrow n=2\)

b^2 = a => b = a/b

bd = 1 => b = 1/d 

=> a/b = 1/d

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

a/b = 1/d = a + 1 / b + d ( đpcm )