Question

Помогите пожалуйста, дам 30 баллов​

Answer 1

Ответ:

1)=(а+b)+(a-b)(a+b)=(a+b)(1+a-b)

2)=(3a-b)²-4²=(3a-b-4)(3a-b+4)

3)=x³(y²-1)-x(y²-1)=(x³-x)(y²-1)=x(x²-1)(y²-1)=

x(x-1)(x+1)(y-1)(y+1)

4)=1-(x²-4xy+4y²)=1-(x-2y)²=(1-x+2y)(1+x-2y)

2x³-50x=0

2x(x²-25)=0

2x=0     x²-25=0

x=0        x²=25

             x=±5

2) 16x³+8x²+x=0

   x(16x²+8x+1)=0

   x(4x+1)²=0

x=0       4x+1=0

             4x=-1

             x=-1/4

3) x³+2x²-36x-72=0

x²(x+2)-36(x+2)=0

(x²-36)(x+2)=0

(x-6)(x+6)(x+2)=0

x-6=0      x+6=0     x+2=0

x=6          x=-6        x=-2

Объяснение:

Answer 2

$\displaystyle\bf\\1)\\\\a+b+a^{2} -b^{2} =(a+b)+(a^{2} -b^{2} )=(a+b)+(a+b)(a-b)=\\\\=(a+b)(1+a-b)\\\\2)\\\\9a^{2}-6ab+b^{2} -16=(9a^{2} -6ab+b^{2} )-16=\\\\=\Big[(3a)^{2} -2\cdot 3a\cdot b+b^{2} \Big]-4^{2} = (3a-b)^{2} -4^{2} =\\\\=(3a-b-4)(3a-b+4)\\\\3)\\\\x^{3} y^{2} -x^{3} -xy^{2}+x=(x^{3} y^{2} -x^{3} )-(xy^{2}-x)=\\\\=x^{3} (y^{2} -1)-x(y^{2}-1)=(y^{2} -1)(x^{3} -x)=\\\\=(y-1)(y+1)x(x^{2} -1)=x(x-1)(x+1)(y-1)(y+1)$

$\displaystyle\bf\\4)\\\\1-x^{2} +4xy-4y^{2} =1-(x^{2} -4xy+4y^{2} )=\\\\=1-\Big[x^{2} -2\cdot x\cdot 2y+(2y)^{2}\Big]= 1-(x-2y)^{2}=(1-x+2y)(1+x-2y) \\\\\\1)\\\\2x^{2} -50x=0\\\\2x(x-25)=0\\\\x_{1} =0\\\\x-25=0 \ \ \Rightarrow \ \ x_{2} =25\\\\2)\\\\16x^{3}+8x^{2} +x=0\\\\x(16x^{2} +8x+1)=0\\\\x_{1} =0\\\\16x^{2} +8x+1=0\\\\D=8^{2} -4\cdot 16 \cdot 1=64-64=0\\\\x_{2} =\frac{-8}{32} =-0,25\\\\3)\\\\x^{3} +2x^{2} -36x-72=0\\\\(x^{3}+2x^{2} )-(36x+72)=0$

$\displaystyle\bf\\x^{2} (x+2)-36(x+2)=0\\\\(x+2)(x^{2} -36)=0\\\\(x+2)(x-6)(x+6)=0\\\\x+2=0 \ \ \Rightarrow \ \ x_{1} =-2\\\\x-6=0 \ \ \Rightarrow \ \ x_{2} =6\\\\x+6=0 \ \ \Rightarrow \ \ x_{3} =-6$