Ta có:\(\left(a-b+c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

      \(=2\left(a-b+c\right)^2-2\left(b-c\right)^2\\ =2\left(\left(a-b+c\right)^2-\left(b-c\right)^2\right)\)

      \(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\\ =2\left(a-2b+2c\right)a \)

\(=2a^2-4ab+4ac\)

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

=-a+b+c+a-b-c=0

hok tốt

Đáp án

= - a + b + c + a - b - c = 0

Hok tốt nhé !

\(-\left(a-b-c\right)-\left(-a+b+c\right)\)

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

\(=0\)

Chúc em học tốt!

1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)

2: \(C=A:B\)

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

\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)

=>C>=-1

a,A= -a-b+c+a+b+c=2c

b, khi a=1,b=-1,c=-2 thì 

A=2(-2)=-4

a)

\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)

\(A=-a-b+c-\left(-a\right)+b+c\)

\(A=-a+\left(-b\right)+c+a+b+c\)

\(A=\left[\left(-a\right)+a\right]+\left[\left(-b\right)+b\right]+\left(c+c\right)\)

\(A=0+0+2c\)

\(A=2c\)

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b)

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

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

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

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

\(A=0+\left(-4\right)=\left(-4\right)\)

Cách 2 : Từ ý   a   suy ra :

\(A=\left(-2\right)\cdot2=\left(-4\right)\)

a) Ta có: -a - b - b = -a - b + c

Vậy: (-a-b+c) - (-a-b-c) = (-a-b+c) - (-a-b+c) = (-a-b+c) : 2

b) (-1-1+-2) : 2 = (-2+-2) : 2 = (-4) : 2 = -2