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1: \(A=\dfrac{15-4+1}{10}+\dfrac{18-8+1}{12}\)
\(=\dfrac{12}{10}+\dfrac{11}{12}\)
\(=\dfrac{6}{5}+\dfrac{11}{12}=\dfrac{72+55}{60}=\dfrac{127}{60}\)
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
\(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-\dfrac{1}{8}-\dfrac{27}{16}}{-\dfrac{29}{16}}\)
\(=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}=1\)
Chúc bạn học tốt!!!
Ta có:
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\\ =\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)
Vậy B = \(\dfrac{72}{5}\)
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}+\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
Đến đây cạn rồi?! ==''
Ta có: \(\left|2x-1\right|-x=4\)
\(\Rightarrow\left|2x-1\right|=4+x\)
+) TH1: \(2x-1\ge0\Rightarrow2x\ge1\Rightarrow x\ge\dfrac{1}{2}\)
Ta có: \(2x-1=4+x\)
\(\Rightarrow2x-x=1+4\)
\(\Rightarrow x=5\) (t/m)
+) TH2: \(2x-1< 0\Rightarrow2x< 1\Rightarrow x< \dfrac{1}{2}\)
Khi đó \(-2x+1=4+x\)
\(\Rightarrow-2x-x=-1+4\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-1\) (t/m)
Vậy \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\).
\(R\left(x\right)=-x^2+\dfrac{2}{3}x+\dfrac{1}{5}\)
\(R\left(x\right)=-1\left(x^2-\dfrac{2}{3}x-\dfrac{1}{5}\right)\)
\(R\left(x\right)=-1\left(x^2-2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^2-\dfrac{1}{5}\right)\)
\(R\left(x\right)=-1\left[\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{45}\right]\)
\(R\left(x\right)=-1\left(x-\dfrac{1}{3}\right)^2+\dfrac{14}{45}\)
\(R\left(x\right)=\dfrac{14}{45}-\left(x-\dfrac{1}{3}\right)^2\le\dfrac{14}{45}\)
Vậy R(x) max = 14/45 tại x = 1/3
1) Vì \(\left|x-2018\right|\) \(\ge\) \(\forall\) x \(\in\) Z
=> \(\left|x-2018\right|+2019\) \(\ge\) 2019
Vậy để biểu thức đạt GTNN \(\Leftrightarrow\)\(\left|x-2018\right|\) = 0
=> x - 2018 = 0
=> x = 0 + 2018
=> x = 2018
Thay x vào biểu thức, ta có:
\(\left|2018-2018\right|\) + 2019
= 0 + 2019
= 2019
R=|2x-4|+|2x+5|+1
=|4-2x|+|2x+5|+1
=>R>=|4-2x+2x+5|+1=10
Dấu = xảy ra khi (2x-4)(2x+5)<=0
=>-5/2<=x<=2
c: Q=|x+1/3|+|2/3-x|>=|x+1/3+2/3-x|=1
Dấu = xảy ra khi (x+1/3)(x-2/3)<=0
=>-1/3<=x<=2/3
\(\dfrac{1}{R\left(x\right)}=\dfrac{1}{x\left(x+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)
\(\Rightarrow S=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{2024}+\dfrac{1}{2023}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2024}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)
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