bài 1

ab+bc+ca=0

=>ab+bc=-ca

ta có (a+b)(b+c)(c+a)/abc

=> (ab+ac+bc+b²)(c+a)/abc

=> (0+b²)(c+a)/abc

=>b²c+b²a/abc

=>b(ab+bc)/abc

=>b(-ac)/abc

=>-abc/abc=-1

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

1)

$2x=a+b+c$

$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$

$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$

$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$

$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$

$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$

$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$

$=ab+bc+ca-x^2$

$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$

2)

$ab+bc+ca=abc,\quad a+b+c=1$

$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$

$\therefore\ (a-1)(b-1)(c-1)=0.$

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

Lời giải:

\(\frac{x-b-c}{a}+\frac{x-a-c}{b}+\frac{x-a-b}{c}=3\)

\(\Leftrightarrow \frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1+\frac{x-a-b}{c}-1=0\)

\(\Leftrightarrow \frac{x-b-c-a}{a}+\frac{x-a-c-b}{b}+\frac{x-a-b-c}{c}=0\)

\(\Leftrightarrow (x-a-b-c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0(1)\)

Vì $abc(ab+bc+ac)\neq 0\Rightarrow \frac{ab+bc+ac}{abc}\neq 0$

$\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq 0(2)$

Từ $(1);(2)\Rightarrow x-a-b-c=0\Rightarrow x=a+b+c$

Cho abc(a+b+c) khác 0. Giải phương trình ẩn x:

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

.

bf

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

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

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

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

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

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

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

Giúp em cái