Tìm x:
a) \(x-\left(\frac{50x}{160}+\frac{25x}{200}\right)=11\frac{1}{4}\)
b)\(\left(x-5\right)\cdot\frac{30}{100}=\frac{200x}{100}+5\)
Giúp mình nhanh với, cảm ơn các bạn rất nhiều!
Những câu hỏi liên quan
Tìm x:
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
Giúp mình nhanh nhé, cảm ơn các bạn rất nhiều!
\(\frac{2}{2.3}\) + \(\frac{2}{3.4}\) + \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + .......+ \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{2019}\) : 2
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2}\) - \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2019}\)
<=> x + 1 = 2019 => x = 2018
vậy x = 2018
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\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
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\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2017}{2019}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
=> x + 1 = 2019
=> x = 2018
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giải phương trình sau
a)\(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
b)\(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
c)\(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
làm giúp mình với mình cần gấp cảm ơn các bạn nhiều
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
tìm x biết rằng :
a) x - \(\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\) c) ( x - 5 ) . \(\frac{30}{100}=\frac{20x}{100}+5\)
a) x-\(\left(\frac{50x}{100}+\frac{25x}{200}\right)\)=\(11\frac{1}{4}\)
<=>x - \(\frac{1}{2}x-\frac{1}{8}x\)=\(\frac{11.4+1}{4}\)
<=>\(\frac{3}{8}x=\frac{45}{4}\)
<=>x=\(\frac{45}{4}:\frac{3}{8}\)
<=>x=30
Vậy x=30
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c) (x-5).\(\frac{30}{100}\)=\(\frac{20x}{100}+5\)
<=>(x-5).\(\frac{3}{10}\)=\(\frac{x}{5}\)+5
<=>\(\frac{3x}{10}-\frac{15}{10}=\frac{x}{5}+5\)
<=>\(\frac{3x}{10}-\frac{x}{5}=5+\frac{15}{10}\)
<=>\(\frac{x}{10}=\frac{13}{2}\)
<=>x=\(\frac{13.10}{2}\)
<=>x=65
Vậy: x=65
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Tìm x, biết:a, x-left(frac{50x}{100}+frac{25x}{200}right)11frac{1}{4} b,left(x-5right).frac{30}{100}frac{200x}{100}+5c,frac{x+1}{2}frac{8}{x+1} d, x:left(9frac{1}{2}-frac{3}{2}right)frac{0,4+frac{2}{9}-frac{2}{11}}{1,6+frac{8}{9}-frac{8}{11}}
Đọc tiếp
Tìm x, biết:
a, \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\) b,\(\left(x-5\right).\frac{30}{100}=\frac{200x}{100}+5\)
c,\(\frac{x+1}{2}=\frac{8}{x+1}\) d, \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)
\(c,\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow(x+1)(x+1)=2.8\)
\(\Rightarrow(x+1)^2=16\)
\(\Rightarrow(x+1)^2=4^2\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
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\(a,x-(\frac{50x}{100}+\frac{25x}{200})=11\frac{1}{4}\)
\(\Rightarrow x-\frac{50x+25x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x}{100}-\frac{75x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x-75x}{100}=\frac{1125}{100}\)
\(\Rightarrow25x=1125\)
\(\Rightarrow x=45\)
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\(b,(x-5)\frac{30}{100}=\frac{200x}{100}+5\)
\(\Rightarrow\frac{30x}{100}-\frac{3}{2}=\frac{200x}{100}+5\)
\(\Rightarrow\frac{30x}{100}-\frac{200x}{100}=5+\frac{3}{2}\)
\(\Rightarrow\frac{-170x}{100}=\frac{13}{2}\)
\(\Rightarrow\frac{-170x}{100}=\frac{650}{100}\)
\(\Rightarrow-170x=650\)
\(\Rightarrow x=\frac{-65}{17}\)
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Chứng minh:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< 1\)
Giúp mình với ạ, cảm ơn các bạn rất nhiều!
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2018}\)= \(\frac{2017}{2018}\)< 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 ( dpcm )
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Ta có:
\(\frac{1}{2^2}\)< \(\frac{1}{1.2}\).
\(\frac{1}{3^2}\)< \(\frac{1}{2.3}\).
\(\frac{1}{4^2}\)< \(\frac{1}{3.4}\).
...
\(\frac{1}{2017^2}\)< \(\frac{1}{2016.2017}\).
\(\frac{1}{2018^2}\)< \(\frac{1}{2017.2018}\).
Từ trên ta có:
\(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+...+ \(\frac{1}{2016.2017}\)+ \(\frac{1}{2017.2018}\)= 1- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{2016}\)- \(\frac{1}{2017}\)+ \(\frac{1}{2017}\)- \(\frac{1}{2018}\)= 1- \(\frac{1}{2018}\)< 1.
=> \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< 1.
=> ĐPCM.
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Tìm x:
a) x - (\(\frac{50x}{100}+\frac{25x}{200}\)) = \(11\frac{1}{4}\)
b) (x - 5) . \(\frac{30}{100}\) = \(\frac{200x}{100}\) + 5
tìm x, biết: \(21+\frac{12}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}=\frac{2011}{2+\frac{3}{4+\frac{5}{6\frac{7}{8}}}}\) ( máy tính casio)
các bạn làm giúp mk bài này với ! cảm ơn nhiều
a) xy-3x-y0b)Làm sao cho frac{10}{x+7}đạt tối giảnc) Cho Afrac{-5x}{21}+frac{-5y}{21}+frac{-5z}{21} biết x+y-zd) Cho Afrac{10^{2004}+1}{10^{2005}+1}vàBfrac{10^{2005}+1}{10^{2006}+1}So sánh A và Be) x-left(frac{50x}{100}+frac{25x}{200}right)11frac{1}{4}f)left(x-5right)timesfrac{30}{100}frac{200x}{100}+5Cái này thực ra là lớp 6 mik ko chọn được Các bạn làm đầy đủ công thức và phép toán nha mình cần gấp
Đọc tiếp
a) xy-3x-y=0
b)Làm sao cho \(\frac{10}{x+7}\)đạt tối giản
c) Cho \(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\) biết x+y=-z
d) Cho \(A=\frac{10^{2004}+1}{10^{2005}+1}\)và\(B=\frac{10^{2005}+1}{10^{2006}+1}\)So sánh A và B
e) \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\)
f)\(\left(x-5\right)\times\frac{30}{100}=\frac{200x}{100}+5\)
Cái này thực ra là lớp 6 mik ko chọn được Các bạn làm đầy đủ công thức và phép toán nha mình cần gấp
Tính
A= \(\frac{1}{31}\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{930}\)
giải nhanh hô mình với, mình cần rất rất rất gấp
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
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