\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

\(=\)\(ab\left(a+b\right)+bc\left(a+b+c-a\right)+ca\left(c+a\right)+2abc\)

\(=\)\(ab\left(a+b\right)+bc\left(a+b\right)+bc\left(c-a\right)+ca\left(c+a\right)+2abc\)

\(=\)\(\left(a+b\right)\left(ab+bc\right)+bc\left(c-a\right)+ca\left(c+a\right)+2abc\)

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

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

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

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

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

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

\(=\)\(b\left(c+a\right)\left(a+b+c\right)+ca\left(c+a\right)\)

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

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

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

...

cách này ngắn hơn nè:

\(ab.\left(a+b\right)+bc.\left(b+c\right)+ac.\left(a+c\right)+2abc\)

\(=a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+abc+abc\)

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

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

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

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

a) \(4\left(x^2-y^2\right)+4x+1\)

\(=4x^2-4y^2+4x+1\)

\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1^2\right]-\left(2y\right)^2\)

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

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

b) \(x^2+1-x^3-x^2\)

\(=1-x^3\)

\(=\left(1-x\right)\left(1+x+x^2\right)\)

\(a^2+b^2+c^2=\left(a+b+c\right)^2-2ab-2bc-2ca=0-2\left(ab+bc+ca\right)=1\)

\(\Leftrightarrow\)\(ab+bc+ca=\frac{-1}{2}\)

\(\Leftrightarrow\)\(\left(ab+bc+ca\right)^2=\left(\frac{-1}{2}\right)^2\)

\(\Leftrightarrow\)\(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=\frac{1}{4}\)

\(\Leftrightarrow\)\(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

\(\Leftrightarrow\)\(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\) ( do \(a+b+c=0\)) 

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

\(M=1^2-2.\frac{1}{4}=1-\frac{1}{2}=\frac{1}{2}\)

... 

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

( x + 2 )² - ( x - 2 )( x + 2 ) = 0

( x + 2 )( x + 2 - x + 2 ) = 0

( x + 2 ) . 4 = 0

x + 2 = 0

x = -2

(x+2)²-(x-2)(x+2)=0

(x+2)(x+2-x+2)=0

(x+2)4=0

=> x+2=0

      x=0-2

      x=-2

                Vậy x=-2

tk cho mk nha!

hình vẽ ( bạn tự vẽ nha)

Vì M,N lần lượt là trung điểm của AB,AC

Nên MN là đường trung bình của \(\Delta ABC\)

=>MN=1/2BC

=>MN=1/2.12

          =6(cm)

               Vậy MN=6cm

tk cho mk nha!

\(x^2-y^2-5x+5y\)

\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)

\(=\left(x+y\right)\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x+y-5\right)\left(x-y\right)\)

1. \(\left(x^2-2x+1\right):\left(x-1\right)+5x=8\)

\(\Rightarrow\left(x-1\right)^2:\left(x-1\right)-5x=8\)

\(\Rightarrow x-1-5x=8\)

\(\Rightarrow-4x-1=8\)

\(\Rightarrow-4x=9\)

\(\Rightarrow x=\frac{-9}{4}\)