\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2017}\left(1+2+...+2017\right)\)
\(=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{2017.2018}{2017.2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2018}{2}\)
\(=\dfrac{1}{2}\left(2+3+...+2018\right)\)
\(=\dfrac{1014045}{2}\)
Thực hiện phép tính
\(a,\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(b,\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(c,\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(d,\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
\(e,\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(f,\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(g,\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(h,\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
Rút gọn:
\(C=\left[\left(\dfrac{1}{a^2+1}\right)\cdot\dfrac{1}{a^2+2a+1}+\dfrac{2}{\left(a+1\right)^3}\cdot\left(\dfrac{1}{a}+1\right)\right]:\dfrac{a-1}{a^3}\)
Bài 1:Giải các pt chứa ẩn ở mẫu sau:
a) \(\dfrac{2x+1}{x-1}=\dfrac{5\left(x-1\right)}{x+1}\) b) \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\) c)\(\dfrac{x-2}{2+x}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
d)\(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\) e)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\) g)\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
h)\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) j)\(\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{50-2x^2}=\dfrac{7}{6\left(x+5\right)}\) k)\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
n)\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)
Rút gọn rồi tính giá trị của biểu thức sau tại \(x=-\dfrac{1}{3}\)
\(\left[\dfrac{x+3}{\left(x-3\right)^2}+\dfrac{6}{x^2-9}-\dfrac{x-3}{\left(x+3\right)^2}\right]\left[1:\left(\dfrac{24x^2}{x^4-81}-\dfrac{12}{x^2+9}\right)\right]\)
Rút gọn:
\(A=\dfrac{x}{5-x}+\left(\dfrac{x}{x^2-25}+\dfrac{5-x}{5x+x^2}\right):\dfrac{2x-5}{x^2+5x}\)
\(B=\left[\left(\dfrac{1}{x^2}+1\right)\cdot\dfrac{1}{1+2x+x^2}+\left(1+\dfrac{1}{x}\right)\cdot\dfrac{2}{\left(1+x\right)^3}\right]:\dfrac{x-1}{x^3}\)
Rút gọn: (sử dụng phương pháp thêm bớt cùng 1 hạng tử)
\(\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)
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Giải các bất phương trình sau rồi biểu diễn tập nghiệm của chúng trên trục số:
1) \(\left(x+3\right)^2-3\left(2x-1\right)>x\left(x-4\right)\)
2) \(1+\dfrac{x+1}{3}>\dfrac{2x-1}{6}-2\)
3) \(x-\dfrac{2x-7}{4}< \dfrac{2x}{3}-\dfrac{2x+3}{2}-1\)
4) \(\dfrac{2x+1}{x-3}\le2\)
5) \(\dfrac{12-3x}{2x+6}>3\)
6) \(x^2+3x-4\le0\)
7) \(\dfrac{5}{5x-1}< \dfrac{-3}{5-3x}\)
8) \(\left(2x-1\right)\left(3-2x\right)\left(1-x\right)>0\)
Rút gọn:
\(A=\left[\dfrac{x+3}{\left(x-3\right)^2}+\dfrac{6}{x^2-9}-\dfrac{x-3}{\left(x+3\right)^2}\right]\left[1:\left(\dfrac{24x^2}{x^4-81}-\dfrac{12}{x^2+9}\right)\right]\)
\(B=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left[\left(x-2\right)+\dfrac{10-x^2}{x+2}\right]\)
Rút gọn biểu thức sau:
\(A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+n}\right)\)với n lá các số tự nhiên lớn hơn 2.
