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bài này khó đấy

Bài này dễ mà

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

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

\(=2n\)

b) Khi \(m=1,n=-1,p=-2\) có :

\(A=2n=2\cdot\left(-1\right)=-2\)

Vậy \(A=-2\) khi \(m=1,n=-1,p=-2\)

Giải:

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

\(A=-m+n-p+m+n+p\) 

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

\(A=0+2n+0\) 

\(A=2n\) 

b) Ta thay: m=1; n=-1; p=-2

Ta có:

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

\(A=\left(-1+-1--2\right)-\left(-1--1--2\right)\) 

\(A=\left(-1-1+2\right)-\left(-1+1+2\right)\) 

\(A=-1-1+2+1-1-2\) 

\(A=\left(-1+1\right)+\left(-1-1\right)+\left(2-2\right)\) 

\(A=0+-2.1+0\) 

\(A=-2\)

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

\(=-m+n-p+m+n+p=2n\)

Vậy A=2n

b) Ta có A=2n

Thay số vào ta được A=2.(-1)=-2

Vậy A=-2 khi n=-1

a)

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

A= -m+n-p+m+n+p

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

A= 0+2n+0

A = 2n

Bài 1: 

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

A = -m + n - p + m + n + p

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

A = 2n

Với n = -1 => A = 2(-1) = -2

Bài 2: 

A = (-2a + 3b - 4c) - (-2a -3b - 4c)

A = -2a + 3b - 4c + 2a + 3b + 4c

A = (-2a + 2a) + (3b + 3b) - (4c - 4c)

A = 6b

Với b = -1 => A = 6(-1) = -6

Bài 3:

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

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

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

A = 2(b - c)

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

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

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

B = 2a

a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)

c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)

\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)

a.    A=-a+b-c+a+b+c

      A=2(b+c)

b.   A=-6

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

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

b: Khi m=1/2 thì \(P=\left(\dfrac{1}{2}-1\right):\dfrac{1}{2}=\dfrac{-1}{2}\cdot2=-1\)

1)

a) A = 2b

b) A = -2

2) a) B = 2n

b) B = -2