\(a)\) ĐKXĐ: \(a\ne-b;a\ne-c;b\ne-c\)

\(\dfrac{x-ab}{a+b}+\dfrac{x-ac}{a+c}+\dfrac{x-bc}{b+c}=a+b+c\)

\(\Leftrightarrow\left(\dfrac{x-ab}{a+b}-c\right)+\left(\dfrac{x-ac}{a+c}-b\right)+\left(\dfrac{x-bc}{b+c}-a\right)=0\)

\(\Leftrightarrow\dfrac{x-ab-ac-bc}{a+b}+\dfrac{x-ac-ab-bc}{a+c}+\dfrac{x-bc-ab-ac}{b+c}=0\)

\(\Leftrightarrow\left(x-ab-ac-bc\right)\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=0\)

Vì \(a,b,c>0\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}>0\)

\(\Leftrightarrow x-ab-ac-bc=0\)

\(\Leftrightarrow x=ab+ac+bc\)

a: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3x^2y+y^3\)

\(=6x^2y+2y^3\)

\(=2y\left(3x^2+y^2\right)\)

a: (x+a)(x+b)

\(=x^2+bx+ax+ab\)

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

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

\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)

\(=x^3+x^2c+ax^2+axc+bx^2+bxc+abx+abc\)

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

Em ko bik ạ

Bài 1:

a ) a.( b² + c² ) + b.( a² + c² ) + c.( a² + b² ) + 2abc

= ab² + ac² + a²b + bc² + a²c + b²c + 2abc

= ( ab² + a²b ) + ( ac² + bc² ) + ( a²c + 2abc + b²c )

= ab.( a + b ) + c².( a + b ) + c.( a² + 2ab + b² )

= ab.( a + b ) + c².( a + b )v + c.( a + b)²

= ( a + b ).[ ( ab + c² + c. ( a + b ) ]

= ( a + b ).( ab + c² + ac + bc )

= ( a + b ).[ ( ab + ac ) + ( c² + bc) ]

= ( a + b ).[ a.( b + c ) + c.( b + c ) ]

= ( a + b ).( b + c ).( a + c )

b) ab.( a + b ) - bc.( b + c ) + ac.( a - c )

= ab.( a + b ) - bc.( b + c ) + ac.[ ( a + b  ) - ( b + c ) ]

= ab.( a + b ) - bc. ( b + c ) + ac.( a + b ) - ac.( b + c )

= ab.( a + b ) + ac.( a + b ) - bc.( b + c ) - ac.( b + c )

= ( a + b ).( ab + ac ) + ( b + c ).( -bc - ac )

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

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

c) ( x² + x )² + 2.( x² + x ) - 3

Đặt x² + x = a

Khi đó đa thức trở thành:

a² + 2a - 3

= a² + 3a - a - 3

= a.( a + 3 ) - ( a + 3 )

= ( a - 1 ).( a - 3 )

\(\Rightarrow\) ( x² + x - 1 ).( x² + x - 3 )

B₂

ab.( a - b ) + bc.( b - c ) + ca.( c - a ) = 0

\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.[ ( a - b ) + ( b - c ) ] = 0

\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.( a - b ) - ca.( b - c ) = 0

\(\Leftrightarrow\)ab.( a - b ) - ca.( a - b ) + bc.( b - c ) - ca.( b - c ) = 0

\(\Leftrightarrow\) ( a - b ).( ab - ca ) + ( b - c ).( bc - ca ) = 0

\(\Leftrightarrow\) ( a - b ).a.( b - c ) - ( b - c ).c.( a - b ) = 0

\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0

\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0

\(\Leftrightarrow\) a = b , b = c , a = c

\(\Rightarrow\) a = b = c

Giải và biện luận các phương trình sau 
a)    (x-ab)/(a+b) + (x-ac)/(a+c) + (x-bc)/(b+c) = a+b+c 

b)    (x-a)/bc + (x-b)/ac + (x-c)/ab = 2(1/a + 1/b + 1/c)

a, Ta có :

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

= -ab + ac  - ad ( ĐPCM)

B,Ta có :

(a+b)(x+y)

= x(a+b) +  y(a+b) ( ĐPCM)