Tính S=\(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+......+\frac{2011}{2011^4+2011^2+1}\)
tính tổng S
S=\(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+...+\frac{2011}{2011^4+2011^2+1}\)
TÍNH : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}}\)
nâng cao phát triển toán 7 đấy
mấy bài đấu thì phải
Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
1. Cho a3+2b2-4b+3=0; a2+a2b2-2b=0
Tính S=a2017+b2016
2. Tính \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+...+\frac{2011}{2011^4+2011^2+1}\)
Giúp với ạ. Cảm ơn nhiều ạ!!!
Câu 1) Ta có\(a^3+2b^2-4b+3=0\Leftrightarrow a^3=-2.\left(b-1\right)^2-1\)\(\le-1\Rightarrow a^3\le-1\Rightarrow a\le-1\Rightarrow a^2\ge1\)
\(\Rightarrow\hept{\begin{cases}a^2\ge1\\a^2b^2\ge b^2\end{cases}}\)\(\Rightarrow a^2+a^2b^2-2b\ge1+b^2-2b\)\(\Leftrightarrow\left(b-1\right)^2\le0\)
Mà \(\left(b-1\right)^2\ge0\)với mọi b nên \(\left(b-1\right)^2=0\)\(\Rightarrow b=1\)
Thay b=1 vào 2 pt ban đầu được \(\hept{\begin{cases}a^3+2-4+3=0\\a^2+a^2-2=0\end{cases}}\)<=> a=1(tm)
Vậy (a,b)=(1;1)
Câu 2 bạn xem ở đây nhé http://olm.vn/hoi-dap/question/716469.html
Tính P:
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
Ta có :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}\right)}\)
\(\frac{1}{2012}\)
Tính P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}\)
\(=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....\left(\frac{1}{2011}+1\right)+1\)
\(=\frac{2012}{2}+\frac{2012}{3}+\frac{2012}{4}+.....+\frac{2012}{2011}+\frac{2012}{2012}\)
\(=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}\right)\)
Thay vào,rút gọn là ra
Tính \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{1010}{2}+\frac{209}{3}+...+\frac{1}{2011}}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)