Question

помогите пж дам 30 балов

Answer 1

(1)

$1) -5a(a^{4} -6a^{2}+3) =-5a^{5} +30a^{3} -15a\\2)(x+4)(3x-2)=3x^{2} -2x+12x-8=3x^{2} +10x-8\\3)(6m+5n)(7m-3n)=42m^{2} -18mn+35mn-15n^{2} =42m^{2}+17mn-15n^{2}\\4)(x+5)(x^{2} +x-6)=x^{3} +x^{2} -6x+5x^{2} +5x-30=x^{3} +6x^{2} -x-30$

(2)

$1)18xy-6x^{2} =6x(3y-x)\\2)15a^{6} -3a^{4} =3a^{4} (5a^{2} -1)\\3)4x-4y+cx-cy=4x+cx-4y-cy=x(4+c)-y(4+c)=(4+c)(x-y)$

(3)

$3x^{2} +9x=0\\3x(x+9)=0\\x=0\\x=-9$

(4)

$7b(2b+3)-(b+6)(b-5)=14b^{2} +21b-b^{2} +5b-6b+30=13b^{2} +20b+30$

(5)

$1)\frac{3x-7}{8}-\frac{x-3}{6} =1\\3(3x-7)-4(x-3)=24\\9x-21-4x+12=24\\5x=24+21-12=33\\x=\frac{33}{5}=6\frac{3}{5} =6.6\\2)(3x+4)(4x-3)-5=(2x+5)(6x-7)\\12x^{2} -9x+16x-12-5=12x^{2} -14x+30x-35\\12x^{2} -12x^{2} -9x+16x-16x =-35+17\\-9x=-18\\x=2$

(6)

$a=0.3\\b=-1\frac{2}{3}$

$24ab+32a-3b-4=24*0.3*(-1\frac{2}{3} )+32*0.3-3*(-1\frac{2}{3} )-4=24*\frac{3}{10} *(-\frac{5}{3}) +9.6-3*(-\frac{5}{3})-4=-24*\frac{1}{2} +9.6+5-4=-12+9.6+5-4=-1.4$