\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

A nguyên khi \(4⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3=\left\{-4;-2;-1;1;2;4\right\}\)

\(\sqrt{x}-3\)	-4	-2	-1	1	2	4
x	\(\varnothing\)	1	4	16	25	49

vậy khi x={1;4;16;25} thì A nguyên

\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\\ =\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\\ =\left(12x^2+11x+0,5+1,5\right)\left(12x^2+11x+0,5-1,5\right)-4\\ =\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\\ =\left(12x^2+11x+0,5\right)^2-6,25\\ =\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\\ =\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

uk, hay lắm, dmm

Ta có 91=7x13.

7, 13 là 2 số nguyên tố nên cần chứng minh B chia hết cho 7 và 13 là nó chia hết cho 91.

ta tìm dư của các số hạng trong B khi chia cho 7 và cộng các dư lại ( chỉ ra tổng các dư đó chia hết cho 7) và suy ra B chia hết cho 7

tương tự với B chia hết cho 13

(cảm ơn bạn đã chia sẻ câu hỏi hay!)

\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)

đẳng thức xảy ra khi :

\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)

vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)

\(A=\dfrac{4}{x}+\dfrac{8}{x\left(x-2\right)}-\dfrac{1}{x-2}\\ A=\dfrac{4\left(x-2\right)+8-x}{x\left(x-2\right)}\\ A=\dfrac{3x}{x\left(x-2\right)}=\dfrac{3}{x-2}\)

A đạt GTLN khi x-2 có GTNN lớn hơn 0

\(x-2>0\Rightarrow x>2\)

vì x là số nguyên nên chọn x=3.

vậy A có GTLN là 3 tại x=3

\(\dfrac{\dfrac{27^3}{81^3}}{3^3}=\dfrac{\dfrac{3^9}{3^{12}}}{3^3}=\dfrac{\dfrac{1}{3^3}}{3^3}=\dfrac{3^3}{3^3}=1\)

\(B=2016^2+2017^2-2\\ B=2016^2-1+2017^2-1\\ B=\left(2016-1\right)\left(2016+1\right)+\left(2017-1\right)\left(2017+1\right)\\ B=2015.2017+2016.2018=A\)

a)

\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\left(đpcm\right)\)

b)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}\)