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a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2x+1).4x=102x−18x.5(2x+1)(2x−1)(2x+1).4x=102x−1
b) (1x2+x−2−xx+1):(1x+x−2)(1x2+x−2−xx+1):(1x+x−2)
=(1x(x+1)+x−2x+1):1+x2−2xx(1x(x+1)+x−2x+1):1+x2−2xx
=1+x(x−2)x(x+1).xx2−2x+11+x(x−2)x(x+1).xx2−2x+1
=(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
=1x−1−x3−xx2+1.[1(x−1)2−1(x−1)(x+1)]
a) (2x+12x−1−2x−12x+1):4x10x−5(2x+12x−1−2x−12x+1):4x10x−5
= 0 - 0
= 0
b) (1x2+x−2−xx+1):(1x+x−2);(1x2+x−2−xx+1):(1x+x−2)
= (x-xx+1) : (2x-2) : (x-xx+1) : (2x-2)
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
= -2x-1-xx2+1. (14 - 4x)
= -x2-1-xx2+14-4x
= -6x-xx2+13
a: \(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x^2-1+x+2-x^2}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x-1}\)
b: P<1
=>P-1<0
=>\(\frac{x^2-x+1}{x-1}<0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được: x<1 và x∉{0;-1}