tìm A biết A =1+1/5+1/5^2+1/5^3+...+1/5^500
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Lời giải:
$A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}$
$5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}$
$\Rightarrow 5A-A=1-\frac{1}{5^{500}}< 1$
Hay $4A< 1$
$\Rightarrow A< \frac{1}{4}$ (đpcm)
a) \(\left(\frac{1}{2}-x\right)^3=\frac{1}{8}\Leftrightarrow\left(\frac{1}{2}-x\right)^3=\left(\frac{1}{2}\right) ^3\)
\(\Leftrightarrow\frac{1}{2}-x=\frac{1}{2}\Leftrightarrow x=\frac{1}{2}-\frac{1}{2}=0\)
b) \(5^{x+1}-5^x=500\Leftrightarrow5^x.5^1-5^x=500\)
\(\Leftrightarrow5^x\left(5-1\right)=500\Leftrightarrow5^x.4=500\)
\(\Leftrightarrow5^x=\frac{500}{4}=125=5^3\).Từ đó ta có: \(5^x=5^3\Leftrightarrow x=3\)
a, (3.x-2)3=2.32
=>(3x-2)3=64
=>(3x-2)3=43
=>3x-2=4
=>3x=6
=>x=2
b,5x+1-5x=500
=>5x(5-1)=500
=>5x*4=500
=>5x=125
=>5x=53
=>x=3
1)
\(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\\ 5A=1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\\ 5A-A=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\right)\\ 4A=1-\dfrac{500}{5^{500}}\\ A=\left(1-\dfrac{500}{5^{500}}\right):4\\ A=1:4-\dfrac{500}{5^{500}}:4\\ A=\dfrac{1}{4}-\dfrac{500}{5^{500}\cdot4}< \dfrac{1}{4}< \dfrac{5}{16}\)
Vậy \(A< \dfrac{5}{16}\)
Bài 1:
Theo đề ra ta có:
$a-2\vdots 3; a-3\vdots 5$
$a-2-2.3\vdots 3; a-3-5\vdots 5$
$\Rightarrow a-8\vdots 3; a-8\vdots 5$
$\Rightarrow a-8=BC(3,5)$
$\Rightarrow a-8\vdots 15$
$\Rightarrow a=15k+8$ với $k$ tự nhiên.
Mà $a$ chia 11 dư 6
$\Rightarrow a-6\vdots 11$
$\Rightarrow 15k+8-6\vdots 11$
$\Rightarrow 15k+2\vdots 11\Rightarrow 4k+2\vdots 11$
$\Rightarrow 4k+2-22\vdots 11\Rightarrow 4k-20\vdots 11$
$\Rightarrow 4(k-5)\vdots 11\Rightarrow k-5\vdots 11$
$\Rightarrow k=11m+5$
Vậy $a=15k+8=15(11m+5)+8=165m+83$ với $m$ tự nhiên.
Vì $a<500\Rightarrow 165m+83<500\Rightarrow m< 2,52$
$\Rightarrow m=0,1,2$
Nếu $m=0$ thì $a=165.0+83=83$
Nếu $m=1$ thì $a=165.1+83=248$
Nếu $m=2$ thì $a=165.2+83=413$
Bài 2:
$a=BC(60,85,90)$
$\Rightarrow a\vdots BCNN(60,85,90)$
$\Rightarrow a\vdots 3060$
Mà $a<1000$ nên $a=0$
Lời giải:
$A=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}$
$5A=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}$
$\Rightarrow 5A-A=5-\frac{1}{5^{500}}$
$\Rightarrow A=\frac{1}{4}(5-\frac{1}{5^{500}})$