+) Xét tử thức: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^2\left(a^2-b^2\right)\)

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

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

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

\(=\left(b-c\right)\left[\left(a^3b-a^2bc\right)+\left(a^3c-a^2c^2\right)+\left(b^2c^2-a^2b^2\right)\right]\)

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

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

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

\(=\left(b-c\right)\left(a-c\right)\left(a-b\right)\left(ab+bc+ca\right)\)

+) Xét mẫu thức: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

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

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

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

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

Từ đó; ta có: 

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

\(=ab+bc+ca\). KL:...

Sửa đề cho nó đẹp

\(\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)

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

em ms hok lớp 1

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

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

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

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

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

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

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

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

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

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .

a.\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ac-2\left(a^2+2ab+b^2\right)=2a^2+2b^2+2c^2+4ab-2a^2-2ab-2b^2=2c^2+2ab\)

b. \(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\left(a^2+b^2-c^2+a^2-b^2+c^2\right)=\left(2b^2-2c^2\right).2a^2=4a^2\left(b^2-c^2\right)=4a^2b^2-4a^2c^2\)

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)

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

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