a, \(\left(-a\right)\left(b-c-d\right)=-ab+ac+ad\)

b, \(x\left(y-z-2\right)=xy-xz-2x\)

c, \(\left(-3\right)\left(2x-5\right)=-6x+15\)

1/\(\left(x-y-z\right)-\left(2x+y+z\right)+\)\(\left(2z-2y\right)=x-y-z-2x-y-z+2z-2y\)

\(\left(x-2x\right)+\left(-y-y-2y\right)+\left(-z-z+2z\right)=-x-4y\)

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

\(=\left(m+m-m\right)+\left(-n-n-n\right)+\left(-p+p\right)=-m-3n\)

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

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

Bài 1:

a. \(=[(3x+(4y-5z)][3x-(4y-5z)]=(3x)^2-(4y-5z)^2\)

\(=9x^2-(16y^2-40yz+25z^2)=9x^2-16y^2+40yz-25z^2\)

b.

\(=(3a-1)^2+2(3a-1)(3a+1)+(3a+1)^2=[(3a-1)+(3a+1)]^2=(6a)^2=36a^2\)

Bài 2:

\((x+y+z)^3=[(x+y)+z]^3=(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3\)

\(=[x^3+y^3+3xy(x+y)]+3(x+y)z(x+y+z)+z^3\)

\(=x^3+y^3+z^3+3xy(x+y)+3(x+y)z(x+y+z)\)

\(=x^3+y^3+z^3+3(x+y)(xy+zx+zy+z^2)\)

\(=x^3+y^3+z^3+3(x+y)(z+x)(z+y)\) (đpcm)

=x+y-z

bài dễ thì nên tự làm, học giỏi hơn

=x+y-z 

 bài toán này rất hay mình vừa học dạng này xong nên mình rất thích cảm ơn bạn

+ (x+y-z) = + x +y -z

\(\text{-(-x-y-z)= x+y+z}\)

\(\text{-(x-y-z)= -x+y+z}\)

\(-\left(-x-y-z\right)=x+y+z\)

\(-\left(x-y-z\right)=-x+y+z\)

Lớp 1 chưa học bài này đâu!!!

x < 3 => x - 3 < 0 => |x - 3| = -(x - 3) = 3 - x

=> M = 3 - x + x - 5 = -2

x < 3 => x - 3 < 0 => (x - 3) = - (x - 3) = 3 - x

=> M = 3 - x + x - 5 = -2

gúp mình với

A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

\(A=x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y\right)\)

\(=x^2-y^2\)

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

\(=\dfrac{1}{4}-4\)

\(=-\dfrac{15}{4}\)