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K
16 tháng 4 2018
Đặt A = \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
2A = \(2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
2A = \(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)
2A + A = \(\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
3A = \(1-\dfrac{1}{64}\)
3A = \(\dfrac{63}{64}\) < 1
hay 3A < 1
=> A < \(\dfrac{1}{3}\)
Vậy .................. (tự kết luận)
NM
20 tháng 5 2017
\(5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{11}{5^{11}}.\)
\(4A=5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}=B-\dfrac{11}{5^{12}}.\)
\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}.\)
\(4B=5B-B=1-\dfrac{1}{5^{11}}\)
\(\Rightarrow4A=\dfrac{1}{4}\left(1-\dfrac{1}{5^{11}}\right)-\dfrac{1}{5^{12}}< \dfrac{1}{4}\Rightarrow A< \dfrac{1}{16}\)
AN
7 tháng 11 2017
a) \(\dfrac{5+x}{4-x}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(5+x\right)=4-x\)
\(\Leftrightarrow2\left(5+x\right)-\left(4-x\right)=0\)
\(\Leftrightarrow10+2x-4+x=0\)
\(\Leftrightarrow6+3x=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
b) \(\dfrac{25}{14}=\dfrac{x+7}{x-4}\)
\(\Leftrightarrow25\left(x-4\right)=14\left(x+7\right)\)
\(\Leftrightarrow25\left(x-4\right)-14\left(x+7\right)=0\)
\(\Leftrightarrow25x-100-14x-98=0\)
\(\Leftrightarrow11x-198=0\)
\(\Leftrightarrow11x=198\)
\(\Leftrightarrow x=18\)
Vậy x=18
c) \(\dfrac{3x-5}{x+4}=\dfrac{5}{2}\)
\(\Leftrightarrow2\left(3x-5\right)=5\left(x+4\right)\)
\(\Leftrightarrow2\left(3x-5\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow6x-10-5x-20=0\)
\(\Leftrightarrow x-30=0\)
\(\Leftrightarrow x=30\)
Vậy x=30
d) \(\dfrac{3x-1}{2x+1}=\dfrac{3}{7}\)
\(\Leftrightarrow7\left(3x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow7\left(3x-1\right)-3\left(2x+1\right)=0\)
\(\Leftrightarrow21x-7-6x-3=0\)
\(\Leftrightarrow15x-10=0\)
\(\Leftrightarrow15x=10\)
\(\Leftrightarrow x=\dfrac{10}{15}=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)
CC
15 tháng 4 2018
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
CC
15 tháng 4 2018
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
14 tháng 3 2018
a,\(\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)
<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)
<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{\left(2x-1\right)\left(1-x\right)}{2\left(1-x\right)}=\dfrac{2\left(1-x\right)}{2\left(1-x\right)}-\dfrac{2\left(x^2-x-3\right)}{2\left(1-x\right)}\)
=>\(5x-2+2x-2x^2-1+x=2-2x-2x^2+2x+6\)
<=>\(-2x^2+8x-3=-2x^2+8\)
<=>\(8x=11< =>x=\dfrac{11}{8}\)
vậy..........
b,\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)
<=>\(\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)
=>\(x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-x+1\)
<=>\(3x^2-25x-6=3x^2-x+1\)
<=>\(-24x=7< =>x=\dfrac{-7}{24}\)
vậy..................
câu c tương tự nhé :)
\(\dfrac{x^2}{16}\)x\(\dfrac{1}{4}\)
= \(\dfrac{x^2}{64}\)
lớp 6 à