\(x^2+x-6\)

\(=x^2+3x-2x-6\)

\(=\left(x^2+3x\right)-\left(2x+6\right)\)

\(=x\left(x+3\right)-2\left(x+3\right)\)

\(=\left(x+3\right)\left(x-2\right)\)

45 = 3².5

840 = 2³.3.5.7

150 = 2.3.5²

900 = 2².3².5²

=> ƯCLN(45, 840, 150, 900) = 3.5 = 15

\(\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)

\(=33-15+20-\left|5\right|-57\)

\(=33-15+20-5-57\)

\(=18+20-5-57\)

\(=38-5-57\)

\(=33-57\)

\(=-24\)

Ta có:

\(45=3^2.5\)

\(60=2^2.3.5\)

\(75=3.5^2\)

\(\Rightarrow BCNN\left(45;60;75\right)=2^2.3^2.5^2=900\)

\(\Rightarrow BC\left(45;60;75\right)=\left\{0;900;1800;2700;3600;4500;5400;6300;........\right\}\)

Ta có:

\(n+5⋮n+1\)

\(\Rightarrow\left(n+1\right)+4⋮n+1\)

\(\Rightarrow4⋮n+1\)

\(\Rightarrow n+1\in U\left(4\right)=\left\{1;2;4\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=2\Rightarrow n=1\\n+1=4\Rightarrow n=3\end{matrix}\right.\)

Vậy \(n\in\left\{0;1;3\right\}\)

a) \(A=34.27+34.73\)

\(\Leftrightarrow A=34.\left(27+73\right)\)

\(\Leftrightarrow A=34.100\)

\(\Leftrightarrow A=3400\)

b) \(B=7^{15}:7^{13}-5^2+2017^0\)

\(\Leftrightarrow B=7^{15-13}-5^2+1\)

\(\Leftrightarrow B=7^2-5^2+1\)

\(\Leftrightarrow B=49-25+1\)

\(\Leftrightarrow B=24+1\)

\(\Leftrightarrow B=25\)

c) \(C=17+\left(-14\right)+\left|-20\right|\)

\(\Leftrightarrow C=17-14+20\)

\(\Leftrightarrow C=3+20\)

\(\Leftrightarrow C=23\)

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