Question

Bài1: cho 2 đa thức A= (1/3a-1/3b)-(a-2b)

                       B= 1/3a-1/3b-(a-b)

Tính A+B và A-B

Bài 2: cho 2 đa thức C= x-[b-(c-a-b)]

                                  D= b+[a-(c-b-a)]

Tính C+D và C-D

 

Answer 1

Bài 1: \(A=\left(\dfrac{1}{3}a-\dfrac{1}{3}b\right)-\left(a-2b\right)\)

\(=\dfrac{1}{3}a-\dfrac{1}{3}b-a+2b=-\dfrac{2}{3}a+\dfrac{5}{3}b\)

\(B=\dfrac{1}{3}a-\dfrac{1}{3}b-\left(a-b\right)\)

\(=\dfrac{1}{3}a-\dfrac{1}{3}b-a+b=-\dfrac{2}{3}a+\dfrac{2}{3}b\)

\(A+B=-\dfrac{2}{3}a+\dfrac{5}{3}b+\dfrac{-2}{3}a+\dfrac{2}{3}b=-\dfrac{4}{3}a+\dfrac{7}{3}b\)

\(A-B=-\dfrac{2}{3}a+\dfrac{5}{3}b+\dfrac{2}{3}a-\dfrac{2}{3}b=b\)

Bài 2:

\(C=x-\left[b-\left(c-a-b\right)\right]\)

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

=x-2b+c-a

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

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

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

C+D

=x-2b+c-a+2a+2b-c

=a+x

C-D

=x-2b+c-a-2a-2b+c

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