Tính
\(\left(\frac{19}{2018}-2019\right)\cdot\frac{1}{2019}-\left(\frac{1}{2018}-2019\right)\cdot\frac{19}{2019}\)
Những câu hỏi liên quan
Giải phương trình:\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}=\frac{19}{49}\)
Đặt \(\left\{{}\begin{matrix}2018-x=a\\x-2019=b\end{matrix}\right.\) \(\Rightarrow a+b=-1\Rightarrow b=-1-a\)
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5a=-3b\\3a=-5b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5a=-3\left(-1-a\right)\\3a=-5\left(-1-a\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=3\\2a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2018-x=\frac{3}{2}\\2018-x=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{4033}{2}\\x=\frac{4041}{2}\end{matrix}\right.\)
tính nhanhfrac{19}{37}+left(1-frac{19}{37}right)frac{7}{13}cdotfrac{5}{14}cdotfrac{39}{15}2frac{3}{7}cdotfrac{1}{2}-frac{1}{2}cdotfrac{3}{7}+frac{1}{3}frac{9}{5}:frac{17}{15}+frac{8}{5}:frac{17}{15}frac{2017}{2018}cdotfrac{1}{2019}+frac{2017}{2018}:frac{2019}{2018}+frac{1}{2018}frac{637cdot527-189}{526cdot637+448}frac{4}{5cdot7}+frac{4}{7cdot9}+frac{4}{9cdot11}+...+frac{4}{23cdot25}dấu . là dấu nhân nha mọi người
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tính nhanh
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(\frac{7}{13}\cdot\frac{5}{14}\cdot\frac{39}{15}\)
\(2\frac{3}{7}\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{3}\)
\(\frac{9}{5}:\frac{17}{15}+\frac{8}{5}:\frac{17}{15}\)
\(\frac{2017}{2018}\cdot\frac{1}{2019}+\frac{2017}{2018}:\frac{2019}{2018}+\frac{1}{2018}\)
\(\frac{637\cdot527-189}{526\cdot637+448}\)
\(\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{23\cdot25}\)
dấu . là dấu nhân nha mọi người
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
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\(\frac{21}{36}\cdot\frac{5}{2}-\frac{7}{12}\cdot\frac{2}{7}+\left(2018-2019\right)^0\)
\(-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
GIUP MINH VOI!!!!!!
\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)
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CÂU 1:
\(a)A=\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{36\cdot\frac{1}{5}\cdot\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(b)B=\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+\frac{9}{39\cdot49}+....+\frac{9}{2009.2019}\)
HELP ME, AI ĐÚNG MÌNH TICK CHO
Tính nhanh :\(\left(3-\frac{1}{18}\cdot\right)\cdot\left(3-\frac{2}{18}\right)\cdot\left(3-\frac{3}{18}\right)\cdot\cdot\cdot\cdot\left(3-\frac{2019}{18}\right)\)
ta có
frac{left(2018-xright)^2+left(2018-xright)left(x-2019right)+left(x-2019right)^2}{left(2018-xright)^2-left(2018-xright)left(x-2019right)-left(x-2019right)^2}frac{19}{49} ( điều kiện : x khác : 2018;2019 )
đặt a x - 2019 ( a khác 0 )
ta có hệ thức :
frac{left(a+1right)^2-left(a+1right)a+a^2}{left(a+1right)^2+left(a+1right)a+a^2}frac{19}{49}
Leftrightarrowfrac{a^2+a+1}{3a^2+3a+1}frac{19}{49}
Leftrightarrow49left(a^2+a+1right)19left(3a^2+3a+1right)
Leftrightarrow49a^2+49a+4957a^2+57a+19...
Đọc tiếp
ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
Tìm MinS= \(\frac{1}{\left(x-2018\right)^2}+\frac{1}{\left(x-2019\right)^2}+\frac{1}{\left(x-2018\right)\left(x-2019\right)}\)
Thực hiện phép tính:
a,\(\left(\frac{9}{16}-\frac{5}{8}+\frac{3}{4}\right):\frac{11}{32}\)
b,\(\frac{1000}{1009}.\frac{-2018}{2019}+\frac{19}{2018}.\frac{-2018}{2019}+\frac{1}{2020}\)
\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)
\(=\frac{11}{16}:\frac{11}{32}\)
\(=\frac{11}{16}.\frac{32}{11}\)
\(=2\)
Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
Tính \(f\left(\frac{1}{2019}\right)+f\left(\frac{2}{2019}\right)+...+f\left(\frac{2018}{2019}\right)\)