\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)

Phân tích mẫu thức thành nhân tử :

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

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

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

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

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

Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)

Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)

Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)

Tử:

(b - c)³ + (c - a)³ + (a - b)³

= (b - c + c - a + a - b)³ - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)

= 0 - 3(b - a)(a - c)(c - b)

= 3(a - b)(a - c)(c - b)

Mẫu:

a²(b - c) + b²(c - a) + c²(a - b)

= a²(b - c) + b²c - ab² + ac² - bc²

= a²(b - c) - a(b² - c²) + bc(b - c)

= a²(b - c) - a(b - c)(b + c) + bc(b - c)

= (b - c)(a² - ab - ac + bc)

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

= (b - c)(a - b)(a - c)

\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)

\(=\frac{3\left(c-b\right)}{b-c}\)

em ms hok lớp 1

Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

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

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

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

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

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

Đặt b - c = x, c - a = y, a - b = z

=> x + y + z = b - c + c - a + a - b = 0

Từ x+y+z=0 => x³+y³+z³=3xyz (tự c/m)

=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)

 đâu khó đâu cái này lớp 6 chứ 8 cái gì

Nếu không khó thì giải giùm đi

Xem lại đề là b²(c² - a²) hay.b⁴(c² - a²​) nhé. Bạn phân tích nhân tử cho tử và mẫu rồi rút gọn là ra nhé. Không khó đâu bạn. Bạn thử làm xem nếu bí quá thì inbox mình chỉ cho

\(BT=\frac{a^2\left(b-c\right)+b^2c-b^2a+c^2a-c^2b}{a^4\left(b^2-c^2\right)+b^4c^2-b^4a^2+c^4a^2-c^4b^2}\)

\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)}{a^4\left(b^2-c^2\right)+b^2c^2\left(b^2-c^2\right)-\left(b^4-c^4\right)a^2}\)

\(=\frac{\left(b-c\right)\left(a^2+bc-a\left(b+c\right)\right)}{\left(b^2-c^2\right)\left(a^4+b^2c^2-a^2\left(b^2+c^2\right)\right)}\)

\(=\frac{\left(a-b\right)\left(a-c\right)}{\left(b+c\right)\left(a^2-b^2\right)\left(a^2-c^2\right)}\)

\(=\frac{1}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)

= \(\frac{a^2\left(b-c\right)+b^2c-c^2b-a\left(b^2-c^2\right)}{a^4\left(b^2-c^2\right)+b^4c^2-c^4b^2-a^2\left(a^4-b^4\right)}\)

= \(\frac{\left(b-c\right)\left(a-b\right)\left(c-a\right)}{\left(b^2-c^2\right)\left(a^2-b^2\right)\left(c^2-a^2\right)}\)

= \(\frac{1}{\left(b+c\right)\left(a+b\right)\left(c+a\right)}\)

\(-\frac{a-c}{a+c}\)