a: \(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)

b: A=1/5

=>\(\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)

=>x^2+1=5x-5

=>x^2-5x+6=0

=>x=2 hoặc x=3

mjk giải baj này mà bận ruj huhu

Giao luu

\(A=\frac{2x\left(x-3\right)+\left(x+3\right)\left(x+1\right)+\left(11x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{2x^2-6x+x^2+4x+3+11x-3}{\left(x+3\right)\left(x-3\right)}=\frac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}=\frac{3x}{x-3}\)

b)\(A=\frac{3x}{x-3}-2< 0\Leftrightarrow\frac{3x-2x+6}{x-3}=\frac{x+6}{x-3}=1+\frac{9}{x-3}\) \(-6< x< 3\)

c) x-3=U(9)=(-9,-3,-1,1,3,9)

x=(-6,0,2,4,6,12)

2/3 + 2/3 = 4/3

3/4 + 5/4 = 8/4 = 2

5/6 + 7/6 = 12/6 = 2

2/9 + 1/18 = 4/18 + 1/18 = 5/18 

2/4 + 1 = 2/4 + 4/4 = 6/4 = 3/2

\(a,ĐK:x\ne1\\ b,A=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\\ c,A=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)

\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)

b: Thay x=-4 vào A, ta được:

\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)

Bạn xem lại đề nhé là \(\sqrt{x-1}\)hay \(\sqrt{x}-1\)

\(\sqrt{x}-1\) mik nhầm đề