先化簡,再求值x28x16x22xx212x

1樓:bc亟

原式=(x?4)

x(x+2)

÷(x+2)(x?2)?12

x+2-1

x+4=(x?4)

x(x+2)

?x+2

(x+4)(x?4)

-1x+4

=x?4?x

x(x+4)

=-4x(x+4)

,不等bai

式組du

?2x>?2

3(x?1)>x?9

,解得:

zhi-3

不等式組的dao整數解為:版-2,-1,0,當x=-1時,原權式=43.

(1)先化簡,再求值:(x-2-12x+2)÷4?xx+2,其中x=-4+3;(2)解方程:xx?1?2x?2x-1=0

2樓:唯愛一萌

(1)原式=(x?2)(x+2)?12

x+2?x+2

4?x=x

?16x+2

?x+2

4?x=-(x+4)(x?4)

x+2?x+2

x?4=-(x+4)

=-x-4,

當x=-4+

3時,回

原式=4-

3-4=-3;

(2)去分母,得:答x2-2(x-1)2-x(x-1)=0,即2x2-5x+2=0,

解得:x1=1

2,x2=2.

經檢驗:x1=1

2,x2=2都是方程的解.

先化簡再求值:1/(x+2)-(x^2+2x+1)/(x-2)÷(x^2-1)/(x-1),其中x=(根號2)-2 5

3樓:匿名使用者

1/(baix+2)du-(x^2+2x+1)/(zhix-2)÷(daox^2-1)/(x-1),其中x=(根號

回2)-2

=1/(x+2)-(x+1)^2/(x-2) *(x-1)/(x-1)(x+1)

=1/(x+2)-(x+1)/(x-2)

=1/(x+2)-(x+1)/(x-2)

=[(x-2-(x+2)(x+1)]/[(x+2)(答x-2)]

=(x-2-x^2-3x-2)/[(x+2)(x-2)]=-(x^2+2x+4)/[(x+2)(x-2)]=-[(√2-2)^2+2(√2-2)+4]/[((√2-2)+2)((√2-2)-2)]

=-[2-4√2+4+2√2-4+4]/[(√2)(√2-4)]=-[6-2√2]/[(√2)(√2-4)]=-[3√2-2]/[(√2-4)]

=-[(3√2-2)(√2+4)]/[(√2+4)(√2-4)]=-[(6+12√2-2√2-8)]/[(2-16)]=-[(-2+10√2)]/[(-14)]=(-1+5√2)/7

4樓:匿名使用者

(4 + 2 x + x^2)/(4 - x^2)

=(3 - sqrt[2])/(-1 + 2 sqrt[2])

5樓:匿名使用者

^解:制1/(x+2)-(x^2+2x+1)/(x-2)÷(x^2-1)/(x-1)

=1/(x+2)-(x+1)2/(x-2) *(x-1)/[(x+1)(x-1)]

=1/(x+2)-(x+1)/(x-2)

=(1-x-1)/(x-2)

=-x/(x-2)

當x=√

2-2時,

原式=-(√2-2)/(√2-2-2)

=(2-√2)(√2+4)/[(√2+4)(√2-4)]=(√2+8-2-4√2)/(2-16)

=(3√2-6)/14

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