x + x / 130 + x / 208 + x / 304 + x / 418 + x/ 550 = 1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
4 tháng 1 2017
\(S=\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}+\frac{1}{x^2+23x+130}+\frac{1}{x^2+29x+208}\)
\(=\frac{1}{x^2+4x+x+4}+\frac{1}{x^2+7x+4x+28}+...+\frac{1}{x^2+16x+13x+208}\)
\(=\frac{1}{x\left(x+4\right)+\left(x+4\right)}+\frac{1}{x\left(x+7\right)+4\left(x+7\right)}+...+\frac{1}{x\left(x+16\right)+13\left(x+16\right)}\)
\(=\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+...+\frac{1}{\left(x+13\right)\left(x+16\right)}\)
\(=\frac{1}{3}\left[\frac{3}{\left(x+1\right)\left(x+4\right)}+\frac{3}{\left(x+4\right)\left(x+7\right)}+...+\frac{3}{\left(x+13\right)\left(x+16\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+...+\frac{1}{x+13}-\frac{1}{x+16}\right]\)
\(=\frac{1}{3}\left[\frac{1}{x+1}-\frac{1}{x+16}\right]\)\(=\frac{1}{3}\left[\frac{x+16}{\left(x+1\right)\left(x+16\right)}-\frac{x+1}{\left(x+1\right)\left(x+16\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{15}{x^2+17x+16}=\frac{5}{x^2+7x+16}\)
1 tháng 3 2022
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
LB
0
ML
4
25 tháng 7 2017
Với sao x + ( x+ 1 ) + ( x + 2 ) Vậy thì các giá trị trong ngoặc sau có x không tkế ???