æ để bài này cho t nhé đợi t thương lượng với chủ thớt r` làm :V
Bài này t làm lần thứ n rồi. Thấy đề là ngán hết muốn làm luôn.
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^{2005}}{b^{2005}} = \dfrac{(a-c)^{2005}}{(b-d)^{2005}}\)
b, \(\dfrac{(a^2+b^2)^3}{(c^2+d^2)^3}\) =\(\dfrac{a^3+b^3)^2}{(c^3+d^3)^2}\)
c, \((\dfrac{a-b}{c-d})^{2005}\) = \(\dfrac{2.a^{2005}-b^{2005}}{2.c^{2005}-d^{2005}}\)
d, \(\dfrac{(a^2-b^2)^5}{(c^2-d^2)^5} = \) \(\dfrac{a^{10}+b^{10}}{c^{10}+d^{10}}\)
e, \(\dfrac{2.a^{2005}+5.b^{2005}}{2.c^{2005}+5.d^{2005}}\) = \(\dfrac{(a+b)^{2005}}{(c+d)^{2005}}\)
f, \(\dfrac{(a^{2004}+b^{2004})^{2005}}{(c^{2004}+d^{2004})^{2005}}\) = \(\dfrac{(a^{2005} -b^{2005})^{2004}}{(c^{2005}-d^{2005})^{2004}}\)
A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)
C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)
E=13+23+...+103=3025
Tính F=23+42+63+...+203=?
Bài 2: Tìm x:
a)\(\dfrac{x-1}{27}\)=\(\dfrac{-3}{1-x}\) c)\(3\times x=2\times y\) và\(x-2\times y=8\)
b)\(\dfrac{4}{5}\)-\(\left|x-\dfrac{1}{2}\right|\)=\(\dfrac{3}{4}\) d)\(\dfrac{x-1}{2005}\)=\(\dfrac{3-y}{2006}\) và x-4009=y
Cho a,b,c. Thỏa mãn: \(\dfrac{a}{2002}=\dfrac{b}{2003}=\dfrac{c}{2005}\)
CMR: \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Chứng minh \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2005^2}< \dfrac{2004}{2005}\)
a,\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+......+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
b,A=\(1+5+5^2+5^3+5^4+.....+5^{49}+5^{50}\)
c,A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
d,A=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
Ai giúp mình thực hiện phép tính này với ạ?? Cảm ơn nhiều!!
A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)
B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
Tính \(\dfrac{B}{A}\)
Cho \(\left(2x_1-3y_1\right)^{2004}+\left(2x_2+3y_2\right)^{2004}+\left(2x_3+3y_3\right)^{2004}+...+\left(2x_{2005}+3y_{2005}\right)^{2004}\le0\)
Chứng minh rằng: \(\dfrac{x_1+x_2+x_3+...+x_{2005}}{y_1+y_2+y_3+...+y_{2005}}=1,5\)
Với a,b,c là số dương chứng minh rằng :
a, \(\left(a+b\right)\times\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
b, \(\left(a+b+c\right)\times\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
