Question

初二几道数学分式题：已知f(x)=x+1／x，求f（√2 - 1）的值。

（1）已知f(x)=x+1／x，求f（√2 - 1）的值。
（分式的乘除法）
（2）(2x-6／4-4x+x^2)÷(x+3)×(x^2+x-6／3-x);（分式的乘除法）
（3）8y^(n-1) × (x^2／n)^n ÷ (- 2x^n／y)^2;

Answer 1

f(√2-1)=√2-1+1/(√2-1)=√2-1+(√2+1)=2√2

[(2x-6)/(4-4x+x^2)]÷(x+3)×[(x^2+x-6)/(3-x)]
=[2(x-3)/[-(x-2)^2]÷(x+3)×[(x+3)(x-2)/[-(x-3)]
=2(x-3)(x+3)(x-2)/[(x-2)^2(x+3)(x-3)]
=2/(x-2)

8y^(n-1) × (x^2／n)^n ÷ (- 2x^n／y)^2;
中间是不是(x^2/y)^n？
=8y^(n-1) × x^2n／y^n ÷ (4x^2n／y^2);
=8y^(n-1)*x^2n*y^2／(y^n*4x^2n)
=(8/4)*y^(n-1+2-n)*x^(2n-2n)
=2y

Answer 2

111

Answer 3

看不懂

Answer 4

f(√2-1)=√2-1+1/(√2-1)=√2-1+(√2+1)=2√2

[(2x-6)/(4-4x+x^2)]÷(x+3)×[(x^2+x-6)/(3-x)]
=[2(x-3)/[-(x-2)^2]÷(x+3)×[(x+3)(x-2)/[-(x-3)]
=2(x-3)(x+3)(x-2)/[(x-2)^2(x+3)(x-3)]
=2/(x-2)

8y^(n-1) × (x^2／n)^n ÷ (- 2x^n／y)^2;
(x^2/y)^n
=8y^(n-1) × x^2n／y^n ÷ (4x^2n／y^2);
=8y^(n-1)*x^2n*y^2／(y^n*4x^2n)
=(8/4)*y^(n-1+2-n)*x^(2n-2n)
=2y