tính tổng sau bằng cách hợp lý
\(B=\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+...+\frac{a}{x^2+19ax+90a^2}+\frac{1}{x+10a}\)
Thực hiện phép tính :
a)\(\frac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\frac{2xy^2}{x^4-2x^2y^2+y^4}+\frac{y^2}{\left(x^2-y^2\right)\left(x+y\right)}\)
b)\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^{8+1}}-\frac{16}{x^{16}+1}\)
c)\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
d)\(\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+....+\frac{a}{x^2+19ax+90a^2}+\frac{1}{x+10a}\)
Rút gọn: \(\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+\frac{a}{x^2+7ax+12a^2}+\frac{a}{x+4a}\)
Rút gọn : \(\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+\frac{a}{x^2+7ax+12a^2}\)\(+\frac{a}{x+4a}\)
Rút gọn \(B=\dfrac{a}{x^2+ax}+\dfrac{a}{x^2+3ax+2a^2}+\dfrac{a}{x^2+5ax+6a^2}+\dfrac{a}{x^2+7ax+12a^2}+\dfrac{a}{x^2+9ax+20a^2}\)
\(B=\dfrac{a}{x^2+ax}+\dfrac{a}{x^2+3ax+2a^2}+\dfrac{a}{x^2+5ax+6a^2}+\dfrac{a}{x^2+7ax+12a^2}+\dfrac{a}{x^2+9ax+20a^2}\)
\(=\dfrac{a}{x\left(x+a\right)}+\dfrac{a}{\left(x+a\right)\left(x+2a\right)}+\dfrac{a}{\left(x+2a\right)\left(x+3a\right)}+\dfrac{a}{\left(x+3a\right)\left(x+4a\right)}+\dfrac{a}{\left(x+4a\right)\left(x+5a\right)}\)
\(=\dfrac{5a}{x^2+5ax}\)
Tính:
a, a/ x^2+ax + a/x^2+3ax+2a^2 + a/x^2+5ax+ 6a^2 + a/x^2 + 7ax+12a^2 + 1/x+4a
b, 1/x^2-x+1 - 1/x^2-x+1 - 2x/x^4-x^2+1 + 4x^3/x^8-x^4+1
Thanks các bạn nha!!
Bài 1: Tính các tổng sau một cách hợp lý nhất:
a) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\) b) \(\frac{2016}{1.3}+\frac{2016}{3.5}+...+\frac{2016}{2015.2017}\)
Bài 2: Tính các tổng sau một cách hợp lý nhất:
a) \(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{399}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
c) \(C=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)
Bài 3: Tìm x bt:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
\(=1008\cdot\frac{2016}{2017}\)\(=\frac{290304}{31}\)Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
bài dài nên tôi ko viết lại đề đâu nhé
Bài 1: Tìm a sao cho đa thức A chia hết cho đa thức B:
1, \(^{x^4+ax^2+1⋮x^2+2x+1}\)
2, \(A=x^2-ax-5a^2-\frac{1}{4}\)
\(B=x+2a\)
3, \(A=6x^4-13ax^3+13a^2x^2-13a^3x-6a^4+\frac{1}{81}\)
\(B=2x^2-3ax-a^2\)
Thực hiện phép tính
a) \(\frac{x^3}{x+1}+\frac{x^2}{x-1}+\frac{1}{x+1}+\frac{1}{1-x}\)
b) \(\frac{x^3}{x-1}-\frac{x^2}{x+1}-\frac{1}{x-1}+\frac{1}{x+1}\)
c) \(\frac{4-2x+x^2}{2+x}-2-x\)
d) \(\frac{2a^3-2b^3}{3a+3b}\times\frac{6a+6b}{a^2-2ab+b^2}\)
Tìm a để các hàm số \(f\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}+ax+1;g\left(x\right)=\frac{x^3}{3}+x^2+3ax+a\) có các điểm cực trị nằm xen kẽ nhau
\(f'\left(x\right)=x^2+2x+3a;g'\left(x\right)=x^2-x+a\)
Ta cần tìm a sao cho g'(x) có 2 nghiệm phân biệt \(x_1\)<\(x_2\) và f'(x) có 2 nghiệm phân biệt \(x_3\)<\(x_4\) sao cho
\(x_1\) <\(x_3\)<\(x_2\) <\(x_4\) và \(x_3\)<\(x_1\)<\(x_4\) <\(x_2\) => \(\begin{cases}\Delta'_1=1-3a>0;\Delta'_2=1-4a>0\\f'\left(x_1\right)f'\left(x_2\right)<0\end{cases}\)
\(\Leftrightarrow\begin{cases}a<\frac{1}{4}\\f'\left(x_1\right)f'\left(x_2\right)<0\end{cases}\) (*)Ta có : \(f'\left(x_1\right)f'\left(x_2\right)<0\) \(\Leftrightarrow\left[g'\left(x_1\right)+3x_1+2a\right]\left[g'\left(x_2\right)+3x_2+2a\right]<0\) \(\Leftrightarrow\left(3x_1+2a\right)\left(3x_2+2a\right)<0\) \(\Leftrightarrow9x_1x_2+6a\left(x_1+x_2\right)+4a^2=a\left(4a+15\right)<0\) \(\Leftrightarrow-\frac{15}{4}\)<a<0