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\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)
\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\)
\(5M=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(5M=1+\frac{1}{5}+...+\frac{1}{5^{49}}\)
\(5M-M=\left(1+\frac{1}{5}+...+\frac{1}{5^{49}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(4M=1-\frac{1}{5^{50}}\)
\(M=\frac{1-\frac{1}{5^{50}}}{4}< \frac{1}{4}=0,25\)
Đpcm
Bài 4:
x O y z m n
Giải:
Vì Om là tia phân giác của góc xOz nên:
mOz = 1/2.xOz
Vì On là tia phân giác của góc zOy nên:
zOn = 1/2 . zOy
Ta có: xOz + zOy = 180o ( kề bù )
=> 1/2(xOz + zOy) = 1/2 . 180o
=> 1/2.xOz + 1/2.zOy = 90o
=> mOz + zOn = 90o
=> mOn = 90o (đpcm)
Bài 2:
7^6 + 7^5 - 7^4 = 7^4.( 7^2 + 7 - 1 ) = 7^4 . 55 chia hết cho 55
Vậy 7^6 + 7^5 - 7^4 chia hết cho 55
A = 1 + 5 + 5^2 + ... + 5^50
=> 5A = 5 + 5^2 + 5^3 + ... + 5^51
=> 5A - A = ( 5 + 5^2 + 5^3 + ... + 5^51 ) - ( 1 + 5 + 5^2 + ... + 5^50 )
=> 4A = 5^51 - 1
=> A = ( 5^51 - 1 )/4
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
5M - M = \(1-\left(\frac{1}{5}\right)^{50}\)hay 4M = \(1-\left(\frac{1}{5}\right)^{50}\)< 1
\(\Rightarrow M=\frac{1-\left(\frac{1}{5}\right)^{50}}{4}< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)(1)
\(\Rightarrow5M=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)(2)
Lấy (2)-(1) ta có
\(\Rightarrow4M=1-\left(\frac{1}{5}\right)^{50}\)
\(\Rightarrow M=\frac{1-\frac{1}{5^{50}}}{4}\)
Do \(1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow M< \frac{1}{4}\)