a) Biến đổi vế trái ta có:

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

= \(x^2+xb+xa+ab\)

= \(x^2+\left(a+b\right)x+ab=VP\)

Vậy đẳng thức đc CM

b) Biến đổi VT ta có:

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

= \(\left(x^2+xa+xb+ab\right)\left(x+c\right)\)

= \(x^3+x^2a+x^2b+x^2c+xab+xac+xbc+abc\)

= \(x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)= VP

Vậy đẳng thức đc CM

2 cái đó chả phải HĐT ai cũng biết hết

Có 2 cách

C1:VT nhân ra

C2:phân tích đa thúc thành nhân tử ở VP

a)  ta có: \(\left(x+a\right)\left(x+b\right)\)

\(=x^2+xb+xa+ab\)

\(=x^2+\left(xb+xa\right)+ab\)

\(=x^2+\left(a+b\right)x+ab\left(ĐPCM\right)\)

Câu b) làm tương tự

HOK TOT

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

\(=x^2+xb+xa+ab\)

\(=x^2+\left(xb+xa\right)+ab\)

\(=x^2+\left(a+b\right)x+ab\left(ĐPCM\right)\)

b, mk chịu

\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)    là Vế Phải 

\(ab+bc+ca-x^2\)là vế trái .

Biến đổi  VP ta có :

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

\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ab\)

\(=3x^2-2x\left(a+b+c\right)+\left(ab+bc+ca\right)\)

Thay \(a+b+c\)là \(2x\)ta được :

\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)= VP

\(=-x^2+ab+bc+ca=VT\)

=> đpcm

- 

a) Sửa đề: \(\left(ax+by+cx\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2\)
= a²x² + b²y² + c²x² + 2axby + 2bycz + 2axcz + b²x² - 2bxay + a²y² + c²y² - 2cybz + b²z² + a²z² - 2azcx + c²x²
= a²x² + b²y² + c²x² + b²x² + a²y² + c²y² + b²z² + a²z² + c²x²
= a²(x²+y²+z²) + b²(x²+y²+z²) + c²(x²+y²+z²)
= (a²+b²+c²)(x²+y²+z²) (đpcm)

b) Đặt x = b; y = c; z = a, ta có:
\(\left(ay+bz+cx\right)^2+\left(az-by\right)^2+\left(bx-cz\right)^2+\left(cy-ax\right)^2\)
= a²y² + b²z² + c²x² + 2aybz + 2bzcx + 2aycx + a²z² - 2azby + b²y² + b²x² - 2bxcz + c²z² + c²y² - 2cyax + a²x²
= a²y² + b²z² + c²x² + a²z² + b²y² + b²x² + c²z² + c²y² + a²x²
= (a²+b²+c²)(x²+y²+z²)
Thay b = x, c = y, a = z, ta có:
(a²+b²+c²)(x²+y²+z²) = (a²+b²+c²)² (đpcm)

a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)