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a) \(-2011-\left(200-2011\right)\)
\(=-2011-200+2011\)
\(=\left(-2011+2011\right)-200\)
\(=0-200\)
\(=-200\)
b) \(\left(-2\right)^2-\left(-2000\right)^0+\left(-1\right)^{2018}-\left|-20\right|\)
\(=4-1+1-20\)
\(=4-20\)
\(=-16\)
Bài 1 :
\(a)-2011-(200-2011)\)
\(=-2011-(200+2011)\)
\(=(-2011+2011)-200\)
\(=0-200=-200\)
\(b)(-2)^2-(-2000)^0+(-1)^{2018}-\left|-20\right|\)
\(=4-1+1-20\)
\(=4-20=-16\)
\(c)23\cdot18-23\cdot26+(-23)\cdot2\)
\(=23\cdot(18-26)+-(23\cdot2)\)
\(=23\cdot(-8)+(-46)\)
\(=-230\)
Bài 2 : Tìm số nguyên x biết :
\(a)3x-(-5)=20\)
\(\Rightarrow3x+5=20\)
\(\Rightarrow3x=20-5\)
\(\Rightarrow3x=15\Rightarrow x=5\)
\(b)3(x+2)=-4+(-2)^3\)
\(\Rightarrow3(x+2)=-4+(-8)\)
\(\Rightarrow3(x+2)=-12\)
\(\Rightarrow x+2=-12\div3\)
\(\Rightarrow x+2=-4\)
Tự tìm x câu b, và câu c,
Bài 3 tự làm
Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
Bài 1:
1. 36.7 + 34.37 + 19.100
= 34.(32.7 + 37) + 19.100
= 81.100 + 19.100
= 100.(81 + 19)
= 100.100
= 10000
2) 2.14.98+7.4.32-28.30
= 28.98 + 28.32 - 28.30
= 28. (98 + 32 - 30)
= 28.100
= 2800
3) (56.35+56.18):53
= [56.(35 + 18)] : 53
= 56.53:53
= 56
4) (158.129-158.39):28
= [158. (129 - 39)] : 28
= 158.90:28
= 5,675
Bài 1:
a) Ta có: \(\left|x-4\right|=\left|-81\right|\)
\(\Leftrightarrow\left|x-4\right|=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=81\\x-4=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=85\\x=-77\end{matrix}\right.\)
Vậy: \(x\in\left\{-77;85\right\}\)
b) Ta có: \(\left(x-2\right)\left(y+1\right)=23\)
\(\Leftrightarrow x-2;y+1\inƯ\left(23\right)\)
\(\Leftrightarrow x-2;y+1\in\left\{1;23;-1;-23\right\}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=23\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=23\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+1=-23\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-23\\y+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=22\end{matrix}\right.\\\left\{{}\begin{matrix}x=25\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-24\end{matrix}\right.\\\left\{{}\begin{matrix}x=-21\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)\(\in\){(3;22);(25;0);(1;-24);(-21;-2)}