ПОЖАЛУЙСТО РЕШИТЕ, УМОЛЯЮ

Задание второе [latex]25a^2-10a+1=(5a-1)^2\\ 64x^2-9y^2=(8x-3y)(8x+3y)\\ 64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)[/latex] [latex]b^5-b^3+b^2-1=b^3(b^2-1)+b^2-1=(b^2-1)(b^3+1)=\\ =(b-1)(b+1)^2(b^2-b+1)[/latex] Задание третье [latex] \dfrac{43^2-17^2}{69^2-9^2} = \dfrac{(43-17)(43+17)}{(69-9)(69+9)} = \dfrac{26\cdot 60}{60\cdot 78} = \dfrac{1}{3} [/latex] [latex] \dfrac{14^2+28\cdot 16+16^2}{15\cdot 17+15\cdot13} = \dfrac{14^2+2\cdot14\cdot 16+16^2}{15\cdot(17+13)} = \dfrac{(14+16)^2}{15\cdot30} = \dfrac{30^2}{15\cdot30} =2[/latex]

2) a) [latex]25a^2-10a+1=(5a-1)^2=(5a-1)(5a-1)[/latex] б) [latex]64x^2-9y^2=(8x)^2-(3y)^2=(8x-3y)(8x+3y)[/latex] в) [latex]64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)[/latex] г) [latex]b^5-b^3+b^2-1=(b^5-b^3)+(b^2-1)=b^3(b^2-1)+(b^2-1)=[/latex][latex]=(b^2-1)(b^3+1)=(b-1)(b+1)(b+1)(b^2-b+1)=[/latex][latex]=(b-1)(b+1)^2(b^2-b+1)[/latex] 3) а) [latex] \frac{43^2-17^2}{69^2-9^2} = \frac{(43-17)(43+17)}{(69-9)(69+9)} = \frac{26*60}{60*78} = \frac{26}{78}= \frac{1}{3} [/latex] б) [latex] \frac{14^2+28*16+16^2}{15*17+15*13} = \frac{14^2+2*14*16+16^2}{15(17+13)} = \frac{(14+16)^2}{15*30} = \frac{30^2}{15*30} = \frac{30}{15} =2[/latex]