giả sử a/2002 = b/2003 = c/2004 = k
=> a = 2002k ; b=2003k và c=2004k
=> 4(a-b)(b-c) = 4(2002k - 2003k)(2003k - 2004k)
=> 2(a-b)(b-c) = 4k^2 (1)
Ta có (c-a)^2 = (2004k - 2002k)^2 = 4k^2 (2)
từ (1) và (2) ta có 2(a-b)(b-c) = (c-a)^2
\(Cho\dfrac{a}{3}=\dfrac{c}{4}=\dfrac{b}{5},CMR:4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
a, cho a,b,c \(\in\) R và a,b,c \(\ne\) 0 thỏa mãn \(b^2=ac\) . CMR : \(\dfrac{a}{c}=\dfrac{\left(a+2013b\right)^2}{\left(b+2013c\right)^2}\)
b, cho cá số a,b,c khác 0 thỏa mãn \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Tính M=\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
Bài 17: Cho a, b, c là 3 số thực khác 0, thỏa mãn điều kiện : \(a+b\ne-c\) và \(\dfrac{a+b-c}{c}\)=\(\dfrac{b+c-a}{a}\)=\(\dfrac{c+a-b}{b}\). Tính giá trị biểu thức P=\(\left(1+\dfrac{b}{a}\right)\)x\(\left(1+\dfrac{a}{c}\right)\)x\(\left(1+\dfrac{c}{b}\right)\)
Câu 1:
a, Tính M =\(3\dfrac{1}{417}\cdot\dfrac{1}{762}-\dfrac{1}{139}\cdot4\dfrac{761}{762}-\dfrac{4}{417\cdot762}+\dfrac{5}{139}\)
b, Tính \(\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\left(\dfrac{3^3}{6}-81\right)...\left(\dfrac{3^{2000}}{2003}-81\right)\)
Câu 2: Cho \(\left(a+3\right)\left(b-4\right)-\left(a-3\right)\left(b+4\right)=0\) . Chứng minh \(\dfrac{a}{3}=\dfrac{b}{4}\).
Cho a, b, c thỏa mãn:
\(\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-a\right)\left(b-c\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}=2013\)
Tính giá trị của biểu thức:
\(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\)
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\) chứng minh \(\dfrac{1}{a^{2005}}+\dfrac{1}{b^{2005}}+\dfrac{1}{c^{2005}}=\dfrac{1}{a^{2005}+b^{2005}+c^{2005}}\)
Cho a,b,c thỏa mãn: \(\dfrac{1}{a+b+c}=\dfrac{a+4b-c}{c}=\dfrac{b+4c-a}{a}=\dfrac{c+4a-b}{b}\)
Tính P= \(\left(2+\dfrac{a}{b}\right).\left(3+\dfrac{b}{c}\right).\left(4+\dfrac{c}{a}\right)\)
Cho 3 số a, b, c thỏa mãn : \(\dfrac{a}{2013}=\dfrac{b}{2014}=\dfrac{c}{2015}\)
Chứng minh \(4\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
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=?