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\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\Rightarrow\frac{x-2}{5x+15}=\frac{303}{1540}\)
\(\Rightarrow x=305\)
ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0
khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)
*ĐK : 101x\(\ge\) 0 => x\(\ge\)0
=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)
\(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)
...
\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)
<=> \(100x+\frac{5050}{101}=101x\)
<=> \(100x-101x=\frac{-5050}{101}\)
<=> -x = -50
=> x = 50 ( t/m x >/ 0)
\(f\left(x\right)=x^{10}+101x^9+101x^8-101x^7+...-101x+101\)
\(=x^{10}-100x^9-x^9+100x^8+x^8-100x^7-x^7+....-101x+101\)
\(=x^9.\left(x-100\right)-x^8\left(x-100\right)+x^7\left(x-100\right)-.....+x\left(x-100\right)-\left(x-101\right)\)
\(\Rightarrow f\left(100\right)=1\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
