a) Tính \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2017}\)
b) So \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\) với \(1\)
So: \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)với 1
giải phương trình
a,(\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\))(x-1)+\(\frac{1}{10}x\)=\(x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-10)+(x-19)+(x-18)+...+100+101=101
d,(\(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+...+\frac{1}{10\cdot60}\))x=\(\frac{1}{1\cdot11}+\frac{1}{1\cdot12}+\frac{1}{1\cdot13}\)
giải phương trình
a,\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-20)+(x-19)+(x-18)+...+100+101=101
a) So sánh : \(\sqrt{17}+\sqrt{26}+1va`\sqrt{99}\)
b) Chứng minh \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Rút gọn B = \(\left[\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right]:\frac{2016}{2017}\)
Tính A=\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}{ }}\)
\(\text{Giải phương trình:}\)
\(a,\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(b,\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
\(c,\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+3}\)
giúp mình với
làm phép tính
c) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)
d) x - 2 \(-\frac{x^2-10}{x+2}\)
e)\(\frac{1}{2x-2y}-\frac{1}{2x+2y}+\frac{y}{y^2-x^2}\)
g)\(\frac{4-2x+x^2}{x+2}-2-x\)
i)\(\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2-x-3}\)