\(2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{68}{288}}}\)
\(2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{68}{288}}}\)
\(2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{68}{288}}}\)
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Những câu hỏi liên quan
5 tháng 10 2021
a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)
HD
3 tháng 9 2017
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
21 tháng 6 2017
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
RM
21 tháng 6 2017
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
KS
7 tháng 5 2017
2155-(174+2155)+(-68+174)=2155-174-2155-68+174
= -68
( 1 - \(\dfrac{1}{2}\) ) ( 1- \(\dfrac{1}{3}\)) ( 1 - \(\dfrac{1}{4}\)) ( 1 - \(\dfrac{1}{5}\)) = \(\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}.\dfrac{1}{5}\)
= \(\dfrac{1}{120}\)
Mình ps có 2 câu à ^.^!
14 tháng 1 2023
a: \(\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^2}=\dfrac{1}{x}\)
=>x^2=x(x^2-x+1)
=>x(x-x^2+x-1)=0
=>x(-x^2+2x-1)=0
=>x=0(loại) hoặc x=1(nhận)
b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2
=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)
=>\(x\simeq28,4\)
WD
2
31 tháng 8 2021
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
Ta có: \(x^2y^2z^2=288^2\)
\(\Leftrightarrow36k^2=288^2\)
\(\Leftrightarrow k^2=2304\)
Trường hợp 1: k=48
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=96\\y=3k=144\\z=6k=288\end{matrix}\right.\)
Trường hợp 2: k=-48
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-96\\y=3k=-144\\z=6k=-288\end{matrix}\right.\)
31 tháng 8 2021
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
\(\Rightarrow4k^29k^236k^2=288^2\)
\(\Rightarrow k^6=64\)
\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4,y=6,z=12\\x=-4,y=-6,z=-12\end{matrix}\right.\)
20 tháng 6 2017
e) \(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{68}}=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}{\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}=\dfrac{1}{3}:\dfrac{1}{4}=\dfrac{3}{4}\)
HH
18 tháng 6 2018
Giải:
h) \(\left(-\dfrac{1}{2}\right)^2.13\dfrac{9}{11}-25\%.6\dfrac{2}{11}\)
\(=\dfrac{1}{4}.\dfrac{152}{11}-\dfrac{1}{4}.\dfrac{68}{11}\)
\(=\dfrac{1}{4}\left(\dfrac{152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}.\dfrac{84}{11}=\dfrac{21}{11}\)
Vậy ...
i) \(57\%+1\dfrac{2}{15}:\dfrac{68}{25}-2,82\)
\(=\dfrac{57}{100}+\dfrac{17}{15}:\dfrac{68}{25}-\dfrac{141}{50}\)
\(=\dfrac{57}{100}+\dfrac{2}{15}-\dfrac{141}{50}\)
\(=\dfrac{-127}{60}\)
Vậy ...
DN
18 tháng 6 2018
h. \(\left(-\dfrac{1}{2}\right)^2\times13\dfrac{9}{11}-25\%\times6\dfrac{2}{11}\)
= \(\dfrac{1}{4}\times\dfrac{150}{11}-\dfrac{1}{4}\times34\)
= \(\dfrac{75}{22}-\dfrac{17}{2}\)
= \(\dfrac{75}{22}-\dfrac{187}{22}\)
= \(-\dfrac{56}{11}\)
i. \(57\%+1\dfrac{2}{15}\div\dfrac{68}{25}-2,82\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\div\dfrac{68}{25}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\times\dfrac{25}{68}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{5}{3}-\dfrac{282}{100}\)
=\(\dfrac{5}{3}-\left(\dfrac{57}{100}-\dfrac{282}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{215}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{43}{20}\right)\)
= \(\dfrac{5}{3}+\dfrac{43}{30}\)
= \(\dfrac{50}{30}+\dfrac{43}{30}\)
= \(\dfrac{31}{10}\)
Bạn không nói gì, mình sẽ rút gọn nhé.
\(2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{68}{288}}}\\ =2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{17}{72}}}\\ =2+\dfrac{1}{2+\dfrac{1}{\dfrac{89}{72}}}\\ =2+\dfrac{1}{2+\dfrac{72}{89}}\\ =2+\dfrac{1}{\dfrac{250}{89}}\\ =2+\dfrac{89}{250}=\dfrac{589}{250}\)
Vậy thôi, chúc bạn học tốt nhé
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cảm ơn bạn nhaa ^^