N = a − 1 a − a : a + 1 a = a − 1 a + 1 a a − 1 : a + 1 a = a + 1 a : a + 1 a = a + 1 a ⋅ a a + 1 = a a = a

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

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

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

\(=2n\)

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

Giải nhanh giúp mình với mình đang cần gấp

x<-1=>x+1<=0

|x+1|=-x-1

x<-1=>2-x>1>0

|2-x|=2-x

N=2-x+3(x+1)=2-x+3x+3=2x+5

a) \(\left|x\right|+x\)

Vì \(\left|x\right|\ge0\) nên ta có 3TH:

TH1: \(x>0\)

\(\Rightarrow\left|x\right|+x=2x\)

TH2: \(x=0\)

\(\Rightarrow\left|x\right|+x=0\)

TH3: \(x< 0\)

\(\Rightarrow\left|x\right|+x=0\)

b) \(N=\left|x\right|:x\)

Vì \(\left|x\right|\ge0\) và \(x\ne0\) nên ta có 2TH:

TH1: \(x>0\)

\(\Rightarrow\left|x\right|:x=1\)

TH2: \(x< 0\)

\(\Rightarrow\left|x\right|:x=-1\)

a: \(M=\left|x\right|+x=\left[{}\begin{matrix}x+x=2x\left(x\ge0\right)\\-x+x=0\left(x< 0\right)\end{matrix}\right.\)

b: \(N=\left|x\right|:x=\pm1\)

\(N=\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}-\sqrt{a}\)

\(N=\dfrac{\sqrt{a}\sqrt{a}+3\sqrt{a}}{\sqrt{a}+3}-\sqrt{a}\)

\(N=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-\sqrt{a}\)

\(N=\sqrt{a}-\sqrt{a}\)

\(N=0\)

\(\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\)

\(=\dfrac{a+3\sqrt{a}-\left(a+3\sqrt{a}\right)}{\sqrt{a}+3}\)

\(=\dfrac{a+3\sqrt{a}-a-3\sqrt{a}}{\sqrt{a}+3}\)

\(=\dfrac{0}{\sqrt{a}+3}\)

\(=0\)