Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
Chứng minh rằng: \(\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{\left(a+b+c\right)^{2n+1}}\)
Tính : \(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{4.6}\right)...\left(1+\dfrac{1}{2015.2017}\right):2\)
Đa thức x 2 – 1 được phân tích thành nhân tử là:
A. (x – 1)(x + 1)
B. (x + 1)(x +1)
C. (- x – 1)(x +1)
D. x(x – 1)
x^3/x+1+x^2/x-1+1/x+1+1/x-1
Tìm giá trị nhỏ nhất của :
a,S=x+ 1/x -1
b,S=x+ 1/x-1 -1
c,S=x+ 1/x+1 -1
d,S=x+ 2/2x+1 -1
Rút gọn \(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)
Rút gọn : \(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right).\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)
Bài 2. Rút gọn các biểu thức sau :
A = (x - 3)(x + 7) – (x + 5)(x - 1) B = - 2(2x + 5)2 – (4x + 1)(1 – 4x)
C = x2(x – 4)(x + 4) – (x2 + 1)(x2 - 1) D = (x + 1)(x2 – x + 1) – (x – 1)(x2 + x +1)
E = (x – 1)3 – (x – 1)(x2 + x + 1) – (3x + 1)(1 – 3x)
(3+1)*(3^2+1)*(3^4+1)*(3^8+1)*(3^16+1)*(3^32+1)
a, T= (1-\(\dfrac{1}{2}\))+(1-\(\dfrac{1}{4}\))+(1-\(\dfrac{1}{8}\))+...+(1-\(\dfrac{1}{512}\))+(1-\(\dfrac{1}{1024}\))+(1-\(\dfrac{1}{2048}\))+(1-\(\dfrac{1}{4096}\))
b, 4*5100*(\(\dfrac{1}{5}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{100}}\))+1