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\(-\dfrac{3}{8}-\dfrac{6}{-15}=-\dfrac{3}{8}+\dfrac{6}{15}=\dfrac{1}{40}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
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Do 160 < 161 nên 2¹⁶⁰ < 2¹⁶¹ (1)
Lại có:
2¹⁶⁰ = (2⁴)⁴⁰ = 16⁴⁰
Do 15 < 16 nên 15⁴⁰ < 16⁴⁰ (2)
Từ (1) và (2) 15⁴⁰ < 2¹⁶¹
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1, \(\left(y+7\right)\left(y-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y+7=0\\y-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-7\\y=5\end{matrix}\right.\)
2, \(25-\left(30+x\right)=\left(-24+3\right)\)
\(\Rightarrow25-\left(30+x\right)=-21\)
\(\Rightarrow30+x=25-\left(-21\right)\)
\(\Rightarrow30+x=25+21\)
\(\Rightarrow30+x=46\)
\(\Rightarrow x=46-30\)
\(\Rightarrow x=16\)
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\(137.\left(-23\right)+137.\left(-17\right)+40.37\\ =137.\left(-23-17\right)+40.37\\ =137.\left(-37\right)+\left(-40\right).\left(-37\right)\\ =\left(137-40\right).\left(-37\right)\\ =-37.97\\ =...\)
Lời giải:
$137(-23)+137(-17)+40.37=137[(-23)+(-17)]+40.37$
$=137(-40)+40.37=-137.40+40.37=40(37-137)=30.(-100)=-3000$
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\(5^3.147-5^3.47+1^{2021}\)
\(=5^3\left(147-47\right)+1\)
\(=125.100+1\)
\(=12500+1\)
\(=12501\)
\(5^3.147-5^3.47+1^{2021}=5^3\left(147-47\right)+1\)
\(=125.100+1=12501\)
A
A
D
C