tinh B=4\(|\)A\(|\)\(+\)\(\frac{1}{3^{100}}\)
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Điều kiện \(\hept{\begin{cases}x-2011>0\\y-2012>0\\z-2013>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2011\\y>2012\\z>2013\end{cases}}}\)
\(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{\sqrt{x-2011}}-\frac{1}{x-2011}+\frac{1}{\sqrt{y-2012}}-\frac{1}{y-2012}+\frac{1}{\sqrt{z-2013}}-\frac{1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2011}-\frac{1}{\sqrt{x-2011}}+\frac{1}{4}\right)+\left(\frac{1}{y-2012}-\frac{1}{\sqrt{y-2012}}+\frac{1}{4}\right)+\left(\frac{1}{z-2013}-\frac{1}{\sqrt{z-2013}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2011}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{y-2012}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{z-2013}}-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2011}}=\frac{1}{4}\\\frac{1}{\sqrt{y-2012}}=\frac{1}{4}\\\frac{1}{\sqrt{z-2013}}=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2011=16\\y-2012=16\\z-2013=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=2027\\y=2028\\z=2029\end{cases}}}\)
(-6,17 +3+5/9-2-36/97)*(1/3-1/4-1/12)=(-6,17+3+5/9-2-36/97)*(4/12-3/12-1/12)=(-6,17+3+5/9-2-36/97)*0=0
Câu a) Mik chữa lại một chút
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\);.......; \(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Suy ra: \(VT< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy : \(VT+1< 1+1=2\)
a, 45 - 45 . ( -12 + 3 )
= 45 - 45 . (-9)
= 45 -(- 405)
=450
b, 31 . ( -18 ) + 31 . ( - 81 ) - 31
= 31. [ -18 + (-81 ) - 1 ]
= 31. (-100)
= -3100
\(2.A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
=> 2.A - A = \(\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
=> A = \(\left(2+\frac{3}{2^2}-1-\frac{100}{2^{100}}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)\)
A = \(1+\frac{3}{2^2}-\frac{100}{2^{100}}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}=\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)+\frac{2}{2^2}-\frac{100}{2^{100}}\)
Tính B = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
2.B = \(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\) => 2.B - B = \(1+\frac{1}{2}-\frac{1}{2^{99}}\)=> B = \(\frac{3}{2}-\frac{1}{2^{99}}\)
Vậy A = \(\frac{3}{2}-\frac{1}{2^{99}}+\frac{2}{2^2}-\frac{100}{2^{100}}=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}=2=\frac{2^{101}-102}{2^{100}}\)