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два варианта ​

Ответ:

Объяснение:

вариант 1

1)(4x²+3)/(x-5)  + ( 2-3x²)/(x-5)=(x²+5)/(x-5)  OTBET: Б

2) x-5 +  15/(x+5)=(x²-25+15)/(x+5)= (x²-10)/(x+5)   OTBET: В

3) x/(x-5) + x/(x+5)=[x(x+5)+x(x-5)] / (x²-25)=(x²+5x+x²-5x)/(x²-25)=2x²/(x²-25) OTBET: Д

4) x/(x-5)  - 5x/(x²-10x+25)=x/(x-5) – 5x/(x²-5)²= [x(x-5)-5x] /(x-5)²=  

= (x²-5x-5x)/(x-5)²=(x²-10x)/(x-5)²  OTBET: A

вариант 2

1) X-4  + 8/(x+4)=(x²-16+8)/(x+4)=(x²-8)/(x+4) OTBET: Г

2) (8x²+1)/(x-4) – (3-7x²)/(4-x)=(8x²+1)/(x-4) + (3-7x²)/(x-4)=(x²+4)/(x-4)  OTBET: В

3)   x/(x-4) – 4x/(x²-8x+16)=x/(x-4) – 4x/(x-4)²= (x²-4x-4x)/(x-4)²=(x²-8x)/(x-4)²                   OTBET: Д

4)  x/(x+4) + x/(x-4)= [x(x-4)+x(x+4)] / (x²-16)=(x²-4x+x²+4x)/(x²-16) = x²/(x²-16)

OTBET: Б