х²+3х+2ху+6у=(х²+3х)+(2ху+6у)=x·(x+3)+2y·(x+3)=(x+3)·(x+2y)
или
х²+3х+2ху+6у=(х²+2ху)+(3х+6y)=)=x·(x+2y)+3·(x+2y)=(x+2y)·(x+3)
28, -14, 7 ...
b₁=28 b₂=-14 b₃=7
q=b₂/b1=-14/28=-1/2
S=b₁/(1-q)=28/(1-(-1/2)=28/1¹/₂=28/³/₂=56/3=18²/₃.
Ответ: S=18²/₃.
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1)(4×(-4)+5)×(-4-3×5)=-11×(-19)=209
2)5(а+б)=5×1,4=7
3)-7(д-с)=-7×(-2)=14