a. ax+ay+bx+by = a (x+y) + b (x+y) 
                         = ( x+y) . (a+b) 
    thay a+b=-2 x+y=17 ta có 
( x+y) . (a+b) = 17 . (-2) = -34
b. ax-ay+bx-by = a(x-y) + b(x-y)    
                          = (x-y) .(a+b) 
thay a+b=-7 x-y=-1
=> (x-y) .(a+b) = (-1) . (-7) =7

Bài 1:

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

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

\(=a+d\)

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

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

\(=-c+d\)

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

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

\(=2a+c\)

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

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

\(=c\)

Bài 3 :

\(a,15-\left(4-x\right)=6\)

                    \(4-x=15-6\)

                    \(4-x=9\)

                            \(x=4-9\)

                            \(x=-5\)

\(b,-30+\left(25-x\right)=-1\)

                          \(25-x=-1+30\)

                          \(25-x=29\)

                                     \(x=25-29\)

                                     \(x=-4\)

\(c,x-5=-1\)

            \(x=-1+5\)

            \(x=4\)

\(d,x-4=-10\)

            \(x=-10+4\)

            \(x=-6\)

\(e,x+3=-8\)

            \(x=-8-3\)

            \(x=-11\)

\(g,x+6=0\)

            \(x=-6\)

Câu 1:

A, (a+b-c)-(b-c-d)

  = a+b-c-b+c+d

  = a+(b-b)+(c-c)

  = a

B, -(a-b+c)+(a-b+d)

  = -a+b-c+a+b+d

  = (a-a)+(b+b)+d-c

  = 2b+d-c

C, (a+b)-(-a+b-c)

  = a+b+a-b+c

  = (a+a)+(b-b)+c

  = 2a+c

D, -(a+b)+(a+b+c)

  = -a-b+a+b+c

  = (-a+a)+(b-b)+c

  = c

  = 

B=x^3+27y^3+9x^2y+27xy^2 

  =x^3+9x^2y+27xy^2+27y^3

 =(x+3y)^3 ---> sử dụng hằng đẳng thức số 4 bạn nhé 

a, Thay số:

\(x+b+c=2+\left(-5\right)+\left(-42\right)=2-5-42=-45\)

b) thay số:

\(x+b+c=0+\left(-34\right)+\left(-16\right)=0-34-16=-50\)

k mk nha. 

#mon

a) x + b + c

Thay x = 2 ; b =-5 ; c = -42

Ta có

2 + ( -5 ) + ( -42) = -45

b) x + b + c

Thay x = 0 ; b = -34 ; c = -16

Ta có :

0 + ( -34 ) + ( -16 ) = -50

Trời!

Bài này tự tính cx đc mak pn

a)Ta thấy: \(\left|x-5\right|\ge0\Rightarrow A\ge0\)

Dấu "=" xảy ra khi \(x=5\)

Vậy \(Min_A=0\) khi \(x=5\)

b)Ta thấy: \(\left|5+x\right|\ge0\Rightarrow B\ge0\)

Dấu "=" xảy ra khi \(x=-5\)

Vậy \(Min_B=0\) khi \(x=-5\)

c)Ta thấy: \(\left|-x+2\right|\ge0\Rightarrow C\ge0\)

Dấu "=" xảy ra khi \(x=2\)

Vậy \(Min_C=0\) khi \(x=2\)

d)Ta thấy: \(\left|x+1\right|\ge0\Rightarrow D\ge0\)

Dấu "=" xảy ra khi \(x=-1\)

Vậy \(Min_D=0\) khi \(x=-1\)

Có: \(M\left(0\right)=a.0^2+b.0+c=c=0\)

      \(M\left(1\right)=a.1^2+b.1+c=a+b+c=0\)

      \(M\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c=0\)

\(M\left(1\right)-M\left(-1\right)=a+b+c-\left(a-b+c\right)\)

\(=a+b+c-a+b-c=2b=0\)

=> \(b=0\)

=> \(a+b+c=a+0+0=a=0\)

Vậy \(a=b=c=0\)

\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{42+462+55}{77}=\dfrac{559}{77}\)

\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{8}\times\dfrac{3}{12}\times\dfrac{9}{5}=\dfrac{243}{480}=\dfrac{81}{160}\)

\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{8}{28}+2=\dfrac{2}{7}+2=\dfrac{16}{7}\)

\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{49}{15}\)

\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{6}{11}+\dfrac{66}{11}+\dfrac{5}{7}=\dfrac{72}{11}+\dfrac{5}{7}=\dfrac{504}{77}+\dfrac{55}{77}=\dfrac{559}{77}\)

\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{32}:\dfrac{5}{9}=\dfrac{9}{32}\times\dfrac{9}{5}=\dfrac{81}{160}\)

\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{2}{7}+2=\dfrac{2}{7}+\dfrac{14}{7}=\dfrac{16}{7}\)

\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{9}{15}+\dfrac{40}{15}=\dfrac{49}{15}\)