\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Tất cả đều phải tìm điều kiện

\(\left(-m+n-p\right)-\left(-m-n-p\right)\)

\(=-m+n-p+m+n+p\)

\(=\left(-m+m\right)+\left(-p+p\right)+\left(n+n\right)\)

\(=2n\)

Vậy \(\left(-m+n-p\right)-\left(-m-n-p\right)=2n\)

\(\left(-m+n-p\right)-\left(-m-n-p\right)\)

\(=-m+n-p+m+n+p\)

\(=\left(-m+m\right)+\left(-p-p\right)+\left(n+n\right)\)

\(=0+0+\left(n+n\right)\)

\(=0+\left(n+n\right)\)

\(=n+n\)

\(=2n\)

Vậy biểu thức (-m+n-p)-(-m-n-p) =2n

\(A=\sqrt{2}-\sqrt{2}\)

\(A=0\)

\(A=\dfrac{1}{\sqrt{2}}\sqrt{14+2\sqrt{13}}-\dfrac{1}{\sqrt{2}}\sqrt{14-2\sqrt{13}}-\sqrt{2}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{13}+1\right)^2}-\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{13}-1\right)^2}-\sqrt{2}\)

\(=\dfrac{\sqrt{13}+1}{\sqrt{2}}-\dfrac{\sqrt{13}-1}{\sqrt{2}}-\sqrt{2}\)

\(=\dfrac{2}{\sqrt{2}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)

A = (-a + b - c) - (-a - b - c)

= -a + b - c + a + b + c

= (a - a) + (b + b) + (c - c)

= 0 + 2b + 0

= 2b

 A = ( -a + b - c) - (-a - b - c)

= -a + b - c + a + b + c

= 2b

=(2a-3)3+(2a-3)2a+3^2-b

=6a-9+(2a)^2-6a+9-b

=6a-6a+9-9+(2a)^2-b

=(2a)^2-b