Lời giải:

a)
$x(a-b)-x(a+b)=xa-xb-xa-xb=-2xb$

b)

$(a^2-ax+by)-(by-a^2-ax)=a^2-ax+by-by+a^2+ax=2a^2$

c)

$(a-b)(a+b)-(b-a)b=a^2-b^2-(b^2-ab)=a^2-b^2-b^2+ab=a^2-2b^2+ab$

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

\(=ax-bx-ax-bx=-2bx\)

\(b,\left(a^2-ax+by\right)-\left(by-a^2-ax\right)\)

\(=a^2-ax+by-by+a^2+ax=2a^2\)

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

\(=a^2-b^2-b^2+ab\)

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

\(=\left(a-2b\right).a\)

a) (m - n) (m + n) = m² + mn - mn + n² = m² + n²

b) x(a - b) - x(a + b) = ax - bx - ax - bx = -2bx

c) (a² - ax + by) - (by - a² - ax) = a² - ax + by - by + a² + ax = 2a²

d) (a - b) (a + b) - (b - a)b = a² + ab - ab - b² - b² + ab = a² - 2b² + ab

cam on ban

a) (m-n)(m+n)=m² +mn-mn-n² =m² -n²

b) x(a-b)-x(a+b)=ax-bx-ax-bx=-2bx

c) (a² -ax+by)-(by-a² -ax)=a² -ax+by-by+a² +ax=2a²

d) (a-b)(a+b)-(b-a)b=a² +ab-ab-b²-b² +ab =a² -2b² +ab

a)A = (a - b) – (a – b + c)

A=a-b-a+b-c

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

A=-c

b)B = (a + b + c) – (a + b - 5)

B=a+b+c-a-b+5

B=(a-a)+(b-b)+(c+5)

B=c+5

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

A = a - b - a + b - c

A = -c

b) B = (a + b + c) - (a + b - 5)

= a + b + c - a - b + 5

= c + 5

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

    A = (a - b) - (a - b) + c

    A = 0 + c

    A = c

b) B = (a + b + c) - (a + b - 5)

    B = c - (-5)

    B = c + 5

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

\(=a^2-ab+ab-b^2-b^2+ab\)

\(=a^2-b^2-b^2+ab\)

\(=a^2-2b^2+ab\)

\(=a\left(b+1\right)-2b^2\)

\(2x-2y=by+cz-cz-ax=by-ax\)

\(\Rightarrow2x-2y=by-ax\)

\(\Rightarrow2x+ax=2y+by\)

\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)

\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)

\(2z-2y=ax+by-cz-ax=by-cz\)

\(\Rightarrow2z+cz=2y+by\)

\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)

\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)

\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)

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

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

=2b+2(-c)

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

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

=2a

a) a+b-a+b+a-c-a-c

=.........................bạn tự đơn giản biểu thức nhé

b) a+b-c+a-b+c-a-c+a-a+b+c

=.....................................

Bài 1: 

a) Ta có: \(-ax\left(xy^3\right)^2\cdot\left(-by\right)^3\)

\(=-a\cdot x\cdot x^2\cdot y^6\cdot\left(-b\right)^3\cdot y^3\)

\(=abx^3y^9\)

b) Ta có: \(xy\cdot\left(-ax\right)^2\cdot\left(-by\right)^3\)

\(=xy\cdot a^2\cdot x^2\cdot b^3\cdot y^3\)

\(=a^2b^3x^3y^4\)

Bài 2: 

Ta có: \(P\left(x\right)=5x-4x^4+x^6+3-2x^3-7x-x^7+1-2x^6+3x^3+x^7\)

\(=\left(-x^7+x^7\right)+\left(x^6-2x^6\right)-4x^4+\left(-2x^3+3x^3\right)+\left(5x-7x\right)+\left(3+1\right)\)

\(=-x^6-4x^4+x^3-2x+4\)

1. Thu gọn đơn thức sau, cho biết phần hệ số, phần biến, bậc của đơn thức(x,y là biến)

a. -ax(xy³)²(-by)³

b. xy(-ax)²(-by)3    

Giải :

a. \(-ax\left(xy^3\right)^2\left(-by\right)^3=-a\cdot x\cdot x^2\cdot\left(y^3\right)^2\cdot\left(-b\right)^3\cdot y^3=-a\cdot\left(-b\right)^3\cdot\left(x\cdot x^2\right)\cdot\left(y^6\cdot y^3\right)=ab^3x^3y^9\)

Phần hệ số : ab³ . Phần biến x³y⁹

Bậc của đơn thức : 3+9=12

b. \(xy\left(-ax\right)^2\left(-by\right)^3=-a^2b^3x^3y^3\)

Phần hệ số : -a²b³. phần biến x³y³

Bậc : 3+3=6

Phân tích mẫu :

\(M=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

Khai triển các bình phương và gom các nhân tử chung :

\(M=\left(ab+ac\right)x^2+\left(ab+bc\right)y^2+\left(bc+ac\right)z^2-2abxy-2bcxy-2acxy\)

\(=\left[\left(ab+ac\right)x^2+a^2x^2+\left(ab+bc\right)y^2+b^2y^2+\left(bc+ac\right)z^2+c^2z^2\right]-\)\(\left(a^2x^2+b^2y^2+c^2z^2+2ab+2aczx+2bcyz\right)\)

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

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\) ( vì \(ax+by+cz=0\) )

Kết quả :  \(M=\frac{1}{a+b+c},a+b+c\ne0\)