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A= 1+ \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ......+ \(\dfrac{1}{1024}\) giúp mk Quỳnh Phương Nguyễn Ace Legona Nguyễn Huy Tú
K
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4 tháng 8 2017
\(A=\dfrac{x+1}{2x-2}+\dfrac{x^2+3}{2-2x^2}+\dfrac{1}{1-x}-15\)
\(A=\dfrac{x+1}{2.\left(x-1\right)}+\dfrac{x^2+3}{2.\left(1-x\right).\left(1+x\right)}+\dfrac{1}{1-x}-\dfrac{15}{1}\)
\(A=\dfrac{\left(x+1\right)^2-\left(x^2+3\right)-2.\left(x+1\right)-15.2.\left(x-1\right).\left(x+1\right)}{2.\left(x-1\right).\left(x+1\right)}\)
\(A=\dfrac{x^2+2x+1-x^2-3-2x-30.\left(x^2-1\right)}{2.\left(x-1\right).\left(x+1\right)}\)
\(A=\dfrac{-4-30x^2+30}{2.\left(x-1\right).\left(x+1\right)}\)
\(A=\dfrac{26-30x^2}{2.\left(x-1\right).\left(x+1\right)}\)
\(A=\dfrac{2.\left(13-15x^2\right)}{2.\left(x-1\right).\left(x+1\right)}\)
\(A=\dfrac{13-15x^2}{\left(x-1\right).\left(x+1\right)}\)
NB
27 tháng 2 2023
Đặt A=1/2+1/4+1/8+..+1/1024
Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)
Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)
<=>A=1-1/1024
<=>A=1023/1024
Vậy biểu thức đã cho = 1023/1024
KT
31 tháng 3 2017
a) \(\dfrac{6x-1}{2-x}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=x\left(3x-2\right)+1\)
\(\Leftrightarrow x-6x^2+2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(3x^2-25x-6-3x^2+2x-1=0\\ \Leftrightarrow-23x-7=0\\ \Leftrightarrow x=\dfrac{7}{-23}\)
vậy phương trình có tập nghiệm là S={\(\dfrac{7}{-23}\)}
KT
31 tháng 3 2017
b)\(\dfrac{x+2}{x-2}-\dfrac{2}{x\left(x-2\right)}=\dfrac{1}{x}\) (ĐKXĐ: \(x\ne2;0\) )
\(\Leftrightarrow\dfrac{x\left(x+2\right)-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-2=x-2\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-1}
HN
10 tháng 6 2017
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=10-1=9\)
10 tháng 6 2017
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
= \(\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
= \(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}+...+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\)
= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
= \(-1+\sqrt{100}\) = -1+10=9
\(\Rightarrow\) đpcm
P
Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
2
31 tháng 12 2023
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
AH
Akai Haruma
Giáo viên
30 tháng 12 2023
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
NT
2
CS
6 tháng 10 2021
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
15 tháng 10 2017
a)
\(\dfrac{x}{5}=\dfrac{y}{7}=k\\ \Rightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
\(x.y=5k.7k=35k^2=140\\ \Rightarrow k^2=4\Rightarrow k=\pm2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=10\\y=14\end{matrix}\right.\\\left\{{}\begin{matrix}x=-10\\y=-14\end{matrix}\right.\end{matrix}\right.\)
15 tháng 10 2017
b)
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{3x}{15}=\dfrac{2y}{6}=\dfrac{7z}{14}=\dfrac{3x-2y+7z}{15-6+14}=\dfrac{69}{23}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=15\\y=9\\z=6\end{matrix}\right.\)
C
1
Ta có:
\(2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(=2.1+2.\dfrac{1}{2}+2.\dfrac{1}{4}+2.\dfrac{1}{8}+...+2.\dfrac{1}{1024}\)
\(=2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
Ta lại có:
\(A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(=2-\dfrac{1}{1024}=\dfrac{2047}{1024}\)
giỏi ghê ta