\(a,5^{10}:5^8+60:12-10\)

\(=5^2+5-10=25+5-10=20\)

\(b,-17+14-12\)

\(=-3-12=-15\)

Bài 2: 

\(a,x-7=14-3.2^3\)

\(\Leftrightarrow x-7=-10\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

\(b,2^x-2^4.2^7.32=0\)

\(\Leftrightarrow2^x=2^{11}.32\)

\(\Leftrightarrow2^x=2^{11}.2^5\)

\(\Leftrightarrow2^x=2^{16}\)

\(\Leftrightarrow x=16\)

Vậy x = 16

ko giải đc đâu e 

\(2xy-y+6x=8\)

\(\Leftrightarrow2xy-y+6x-8=0\)

\(\Leftrightarrow y\left(2x-1\right)+3.\left(2x-1\right)-5=0\)

\(\Leftrightarrow\left(2x-1\right)\left(y+3\right)=5\)

Đến chỗ này thì xét thuộc U(5) nhé

\(\left(2x+3\right)⋮\left(3x+2\right)\Rightarrow3\left(2x+3\right)=6x+9=6x+4+5=2\left(3x+2\right)+5⋮\left(3x+2\right)\)

\(\Leftrightarrow5⋮\left(3x+2\right)\Leftrightarrow3x+2\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)

\(\Leftrightarrow x\in\left\{-1,1\right\}\)(vì \(x\)nguyên) 

Thử lại đều thỏa mãn. 

\(x\left(1-3y\right)-y=0\Rightarrow x=\frac{y}{1-3y}\)

\(\Rightarrow3x=\frac{3y}{1-3y}=\frac{-\left(1-3y\right)+1}{1-3y}=-1+\frac{1}{1-3y}\)

x nguyên => 3x nguyên => \(\frac{1}{1-3y}\) nguyên hay 1-3y là ước của 1

\(\Rightarrow1-3y=\left\{-1;1\right\}\Rightarrow y=0\Rightarrow x=0\)

( -2 )³ . 5 -( -3) . 7 + ( -10 )⁰

^{=(-8).5-(-21)+1}

^=(-40)+21+1

^=-18

^HT