Question

1)разложите на множители : а)3а²-6а³+18а4(маленькая,сверху) б)7а³в³-77а²в²-21а³в4(маленькая,сверху) в)а*(3в+с)-х*(3в+с) 2)решите систему: а)3а+2в=1 б)х-2 + у-2 2а+3в=8 =2 4 4 х-2 у-2 - = 4/3 3 9

Answer 1

$1) a) 3a^2-6a^3+18a^4=3a^2(1-2a+6a^2) \\ b) 7a^3b^3-77a^2b^2-21a^3b^4=7a^2b^2(ab-11-3ab^2) \\ c) a(3b+c)-x(3b+c)=(a-x)(3b+c) \\$
$2) a) \left \{ {{3a+2b=1} | *2 \atop {2a+3b=8}|*3} \right. \\ -\left \{ {{6a+4b=2} \atop {6a+9b=24}} \right. \\ -5b=-22 \\ 5b=22 \\ b= \frac{22}{5} \\ 3a+2* \frac{22}{5} =1 \\ 3a+ \frac{44}{5} =1 \\ 3a=1- \frac{44}{5} \\ 3a=- \frac{39}{5} \\ a= -\frac{39}{5*3} =- \frac{39}{15} =- \frac{13}{5} \\ OTBET: a=- \frac{13}{5} ; b= \frac{22}{5}$
$b) \left \{ {{ \frac{x-2}{4} + \frac{y-2}{4} =2} \atop { \frac{x-2}{3} - \frac{y-2}{9} = \frac{4}{3} }} \right. \\ \left \{ {{ \frac{x-2+(y-2)}{4} =2} \atop { \frac{3(x-2)-(y-2)}{9} = \frac{12}{9} }} \right. \\ \left \{ {{x-2+y-2=8} \atop {3x-6-y+2=12}} \right. \\ + \left \{ {{x+y=12} \atop {3x-y=16}} \right. \\ 4x=28 \\ x=28:4=7 \\ y=12-x=12-7=5 \\ OTBET: x=7; y=5$