Nếu \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^m+1\right)+1=2^{3m-218}\) thì m bằng ...
K
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KN
2 tháng 11 2019
a) Ta có:
\(\frac{1}{2\left(m+1\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3m+2}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}\)
\(+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3m+3}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3\left(m+1\right)}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3}{2\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3\left(8m+5\right)}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{24m+15}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{24m+16}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{8\left(3m+2\right)}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{8}{2\left(8m+5\right)}=\frac{4}{8m+5}\left(đpcm\right)\)
KN
2 tháng 11 2019
b) Ta có: \(\frac{1}{m+1}+\frac{1}{3m+2}+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{3m+2}{\left(m+1\right)\left(3m+2\right)}+\frac{m+1}{\left(m+1\right)\left(3m+2\right)}\)
\(+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4m+4}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4\left(m+1\right)}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4}{3m+2}\left(đpcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 6 2020
a/ \(\left\{{}\begin{matrix}3m+1>0\\\Delta=\left(3m+1\right)^2-4\left(3m+1\right)\left(m+4\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-\frac{1}{3}\\\left(3m+1\right)\left(-m-15\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-\frac{1}{3}\\\left[{}\begin{matrix}m\ge-\frac{1}{3}\\m\le-15\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>-\frac{1}{3}\)
b/\(\left\{{}\begin{matrix}m+1>0\\\Delta'=\left(m-1\right)^2-\left(m+1\right)\left(3m-3\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-1\\\left(m-1\right)\left(-2m-4\right)\le0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m>-1\\\left[{}\begin{matrix}m\ge1\\m\le-2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m\ge1\)
10 tháng 3 2022
\(f\left(x\right)=\left(3m-4\right)x^2-2\left(m-2\right)x+m-1< 0\)
\(TH1:3m-4=0\Leftrightarrow m=\dfrac{4}{3}\Rightarrow f\left(x\right)=\dfrac{4}{3}x+\dfrac{1}{3}< 0\Leftrightarrow x< -\dfrac{1}{4}\left(ktm\right)\)
\(TH2:3m-4>0\Leftrightarrow m>\dfrac{4}{3}\Rightarrow f\left(x\right)< 0\forall x>1\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\x1\le1< x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)^2-\left(m-1\right)\left(3m-4\right)>0\\\left(x1-1\right)\left(x2-1\right)\le0\Leftrightarrow x1.x2-\left(x1+x2\right)+1\le0\\\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0< m< \dfrac{3}{2}\\\dfrac{m-1}{3m-4}-\dfrac{2\left(m-2\right)}{3m-4}+1\le0\Leftrightarrow\dfrac{1}{2}\le m< \dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\le m< \dfrac{4}{3}\left(màm>\dfrac{4}{3}\right)\Rightarrow loại\)
\(TH3:3m-4< 0\Leftrightarrow m< \dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\Delta'=0\Leftrightarrow m=0\left(tm\right)\\x=\dfrac{2\left(m-2\right)}{3m-4}=\dfrac{1}{2}\notin\left(1;+\infty\right)\left(tm\right)\end{matrix}\right.\\\Delta'< 0\Leftrightarrow\left[{}\begin{matrix}m< 0\\m>\dfrac{3}{2}\end{matrix}\right.\\x1< x2\le1\left(1\right)\\\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\Leftrightarrow0< m< \dfrac{3}{2}\\\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0< m< \dfrac{3}{2}\\\dfrac{m-1}{3m-4}-\dfrac{2\left(m-2\right)}{3m-2}+1\ge0\\\dfrac{2\left(m-2\right)}{3m-4}-2< 0\end{matrix}\right.\)
\(\Leftrightarrow0< m\le\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}m\le0\\0< m\le\dfrac{1}{2}\end{matrix}\right.\)
11 tháng 3 2022
thay \(\dfrac{1}{2}\) vào ra x<1/5 hoặc x>1 chứ có phải Vx>1 đâu ạ
<=> (24-1)(24+1).....(2m+1)=23m-218
<=> 22m-1+1=23m-218
<=> 22m=23m-218
<=>2m=3m-218
=>m=218
<=> (24-1)(24+1).....(2m+1)+1=23m-218
<=> 22m-1+1=23m-218
<=> 22m=23m-218
<=>2m=3m-218
=>m=218
ở dưới mình nhầm nha!!!