50 балл!!! Решите полностью!!! 4- и 5 задание быстро!!!!

4) Пускай  первое число -a , второе -b. [latex]a) \left \{ {{a*b=-32} \atop {a+b=-14}} \right. ; \left \{ {{a=-14-b} \atop {(-14-b)*b=-32}} \right. (-14-b)*b=-32 -14b-b^2=-32 b^2+14b-32=0 D=b^2-4ac=14^2 -4*1*(-32)=324; \sqrt{D}=18 b_{1} = \frac{-b-\sqrt{D}}{2a}= \frac{-14-18}{2}=-16 b_{2}= \frac{-b+\sqrt{D}}{2a}= \frac{-14+18}{2}=2 a_{1} =-14-(-16)=-14+16=2 a_{2} =-14-2=-16[/latex] [latex]b) \left \{ {{a*b=65} \atop {a+b=-18}} \right. ; \left \{ {{a=-18-b} \atop {(-18-b)*b=65}} \right. (-18-b)*b=65 (-18-b)*b=65 -18b-b^2=65 b^2+18b+65=0 D=b^2-4ac=18^2 -4*1*(65)=324-260=64; \sqrt{D}=8 b_{1} = \frac{-b-\sqrt{D}}{2a}= \frac{-18-8}{2}=-13 b_{2} = \frac{-b-\sqrt{D}}{2a}= \frac{-18+8}{2}=-5 a_{1} =-18-(-13)=-18+13=-5 a_{2} =-18+5=-13[/latex] [latex]c) \left \{ {{a*b=72} \atop {a-b=1}} \right. ; \left \{ {{a=1+b} \atop {(1+b)*b=72} \right. (1+b)*b=72 b+b^2=72 b^2+b-72=0 D=b^2-4ac=1^2 +4*1*(72)=289; \sqrt{D}=17 b_{1} = \frac{-b-\sqrt{D}}{2a}= \frac{-1-17}{2}=-9 b_{2} = \frac{-b-\sqrt{D}}{2a}= \frac{-1+17}{2}=8 a_{1} =1+(-9)=-8 a_{2} =1+8=9[/latex] [latex]d) \left \{ {{a+b=-12} \atop {a-b=-2}} \right. ; \left \{ {{a=-12-b} \atop {(-12-b)-b=-2}} \right. (-12-b)-b=-2 -12-b-b=-2 -2b=10 b=-5 a=-12-(-5)=-12+5=-7[/latex] 5) [latex]a) (-4)^2=16; 4^2=16; (-12)^2=144; (-2)^3=-8; 25^2=625; (-100)^3=-1 000 000;(-1)^3=-1;3^3=27;-4^2=-16 b)15^2-10^3=225-1000=775; c)100^2-(-16)^2=10000-256=9744 d)(-5)^3+(-5)^2=-125+25=100 e)(-5)^2-2^3=25-8=17 f)3^2-(5^2-2^3)=9-(25-8)=9-17=-8 g)4^3-(-3)^3=64+27=91 h)(-2)^3*3^2*4=-8*9*4=-288 i)(-2)^2+(-2)^2=4+4=8[/latex]