Grade 8 Math Problem Simplification 1 x x 1 x x 1 x 1 x 1998

You make this polynomial simple to s

s=1+x+x(1+x)+x(1+x)²+x（1+x)^1998

1+x)s=(1+x) +x(1+x) +x(1+x) 1998+x(1+x) 1999, subtract the two formulas, and sort it out, s=(1+x) 1999

First, write the formula under the root sign as (x-1) 2 x 2

Obviously, after the sail is opened, it is the absolute code impulse source value of (x-1) x.

When x>1, the absolute value of x-1 is greater than zero, so the result of this proposal is (x-1) delay state x=1-1 x

1.By |x+1|+|x-1|

1) When x -1: the original formula = -x-1 + 1-x = -2x

2) When -1 x 1: original = x+1+1-x=2, (3) x 1: original = x+1+x-1=2x.

2.（1+a）（1-a）=1-a²。

.Simplify: 丨x+1丨+丨x-1丨=2(-1<=x<=1)丨x+1丨+丨x-1丨=2x (x>1).

丨x+1丨+丨x-1丨=-2x (x<-1)2. (1+a)(1-a)=1-a^2

Solution: 1+x+x(1+x)+x(1+x) 2+.x(1+x)^2002

1+x)+x[(1+x)+(1+x)^2+..1+x)^2002](1+x)+xsn

It is easy to see that this one here is an equal proportional series, according to the proportional series formula.

Because, a1=1+x, q=1+x

So, sn=[a1(1-q n)] 1-q(a1-a1*q n) 1-q

1+x)-(1+x)(1+x) 2002] 1-(1+x)[(1+x)-(1+x) 2003] (x) substituting sn into the above equation, 1+x+x(1+x)+x(1+x) 2+...x(1+x)^2002

1+x)+xsn

1+x)+x[(1+x)-(1+x)^2003]/(x)(1+x)-[1+x)-(1+x)^2003](1+x)^2003

The root number is 1-x>=0,x,<=1

x-1>=0,x.=1

If both are true, then x=1

So the original formula = 0 + 0 = 0

The two radicals (1-x) x-1) make sense 1-x 0, x-1 0

1 x 1, i.e. x = 1

(1-x) + x-1) = 0 + 0 = 0 real number a satisfies | 2010-a | a-2011) =a,a=2010²-2011

a-2012^2=2010²-2011-2012²=-2011×5=-10055

x is greater than or equal to 1 and x is less than or equal to 1 so x is equal to 1 so the original formula is equal to 0

Take the definition field for the sub of the equation and get x=0

So the original = 0