A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{308}\)+ \(\frac{1}{309}\)

B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+ \(\frac{306}{3}\)+\(\frac{3}{306}\) + \(\frac{2}{307}\)+ \(\frac{1}{308}\)

=> B = \(\frac{309-1}{1}\)+ \(\frac{309-3}{3}\)+... + ( 309 ... )

=> B = 309 + 309 . ( \(\frac{1}{2}\) + \(\frac{1}{3}\)+... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\)- \(\frac{1}{1}\)+ \(\frac{2}{2}\)+ ... + \(\frac{308}{308}\)+ \(\frac{309}{309}\)

=> B = 309 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))

=> \(\frac{A}{B}\)= \(\frac{1}{309}\)

Lâu rồi bạn còn cần lời giải ko mình giải cho

Ta có :

\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)

\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)

\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)

\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)

\(\frac{A}{B}=\frac{1}{309}\)

\(x\inℤ\) hả 

x thuộc Z

CHỜ TÍ

\(B=308/1+307/2+306/3+...+1/308 \)

\(B=308+307/2+306/3+...+1/308\) chia số 308 thành 308 số 1

B=307/2+1+306/3+1+...+1/308+1+1

B=309/2+309/3+309/4+...+309/308+309/309

B=309(1/2+1/3+1/4+...+1/309)=309A

Suy ra A/B=1/309

=(1/2+1/31/4...1/307/1/3081/309)/(309-1/1+309-2/2+...+309-307/307+309-308/308)

=(1/21/31/4...1/3071/3081/309)/(309/1-1+309/2-1+...+309/307-1+309/308-1)

=(........................................)/(309/309309/2309/3...309/307+309/308)

=(........................................)/[309x(1/309+1/308+...+1/41/31/2)]

Thấy tử và mẫu giống nhau thì ta rút:

=1/309

      Bài giải

a/b = 2/3 => a²/b² = 2.2/3.3 = 4/9

a² + b² = 208

a² = 208 : (4 + 9).4

a² = 208 : 13.4

a² = 16.4

a² = 64

=> a = 8

=> b = 8 : 2/3 = 12

Ta có \(\frac{a}{b}=\frac{2}{3}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{2}{3}\right)^2\Rightarrow\frac{a^2}{b^2}=\frac{4}{9}\)

Theo tính chất của tỉ lệ thức thì ta có \(\frac{a^2}{4}=\frac{b^2}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có \(\frac{a^2}{4}=\frac{b^2}{9}=\frac{a^2+b^2}{4+9}=\frac{208}{13}=16\)

\(\Rightarrow\hept{\begin{cases}a^2=16.4=64\\b^2=16.9=144\end{cases}}\)

Vì \(\frac{a}{b}=\frac{2}{3}\) nên a, b cùng âm hoặc cùng dương.

Vậy \(\orbr{\begin{cases}a=8,b=12\\a=-8,b=-12\end{cases}}\)

\(b^2hay\)2b vậy

Bài 1: 

\(=\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}+\dfrac{1}{\left(x+13\right)\left(x+16\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+13\right)}+\dfrac{3}{\left(x+13\right)\cdot\left(x+16\right)}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+13}+\dfrac{1}{x+13}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{x+16-x-1}{\left(x+1\right)\left(x+16\right)}=\dfrac{5}{\left(x+1\right)\left(x+16\right)}\)

Bài 2: 

\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+4\right)^2+\left(2c-1\right)^2=0\)

Dấu '=' xảy ra khi a=1; b=-4; c=1/2

22445

208*24=384*(a-2)

4992=384*(a-2)

a-2=4992/384=13

a=13+2=15

208/384=208/384:13=16/24

a + 2 = 16

a        = 16-2

a         =14