Вправи: 233 (1)

233. Доведіть, що значення виразу не залежить від значень

змінної х на її ОДЗ:

1) ((x+2x-1)² + 3) : ((x-1x+1)² + 3) : x3+1x3-1 – 2xx-1 = ((x+1)2(x-1)2 + 3\(x-1)2) : ((x-1)2 (x+1)2 + 3\(x+1)2) : x3+1x3-1 – 2xx-1 = (x+1)2+ 3(x-1)2(x-1)2 : (x-1)2+3(x+1)2 (x+1)2 : x3+1x3-1 – 2xx-1 = x2+2x+1+3(x2-2x+1)(x-1)2 : x2-2+1+3(x2+2x+1)(x+1)2 : x3+1x3-1 – 2xx-1 = x2+2x+1+ 3x2-6x+3(x-1)2 : x2-2x+1+3x2+6x+3 (x+1)2 : x3+1x3-1 – 2xx-1 = 4x2-4x+4(x-1)2 : 4x2+4x+4(x+1)2 :

(x+1)(x2- x+1)(x-1)(x2+ x+1)– 2xx-1= 4(x2- x+1)(x-1)2 • (x+1)24(x2+ x+1) • x-1(x2+ x+1)x+ (x2- x+1) – 2xx-1 = 4(x2- x+1)• (x+1)2• (x-1)(x2+ x+1)(x-1)2• 4(x2+ x+1)(x+1)(x2- x+1) – 2xx-1 = x+1x-1 – 2xx-1 = x+1-2xx-1 = 1- xx-1 = –x-1x-1 = –1;

2) (1+ x1- – 1- x1+ x + 4x2x2-1) : (2-2xx2 + 2 – 2x3+x2) = (–x+1\x+1x-1 + x-1\x-1x+1 + 4x2(x-1)(x+1)) : (2(1- x)\x+1x2 + 2\x2(x+1)1 – 2x2(x+1)) = -(x+1)2+(x-1)2+4x2 (x-1)(x+1) : 2(1- x2)+2x2x+1-2 x2(x+1) = -(x2+2x+1)+x2-2x+1+ 4x2 (x-1)(x+ 1 : 2(1- x2)+ 2x2x+1-2x2(x+1) = -x2-2x-1+x2-2x+1+ 4x2 (x-1)(x+1) : 2-2x2+2x3+2x2-2 x2(x+1) = 4x2- 4x(x-1)(x+1) : 2x3x3(x+1) = 4x(x-1)(x-1)(x+1) • x2(x+1)2x3 = 4xx-1• x2(x+1)x-1x+1• 2x3 = 4x32x3 = 2.