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

\(=x^3+ax^2+bx^2+cx^2+abx+acx+bcx+abc\)

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

\(=x^3+6x^2-7x-60\)

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

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

= \(x^3+x^2c+x^2b+xbc+ax^2+axc+abx+abc\)

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

= \(x^3+6x^2-7x-60\)

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

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

    \(=x^3+ax^2+bx^2+abx+cx^2+acx+bcx+abc\)

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

Theo bài ra ta có:

\(a+b+c=6\)

\(ab+bc+ca=-7\)

\(abc=-60\)

\(\Rightarrow A=x^3+6x^2-7x-60\)

$\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3$

$\Leftrightarrow x\left(\dfrac1a+\dfrac1b+\dfrac1c\right)-\left(\dfrac ba+\dfrac ca+\dfrac cb+\dfrac ab+\dfrac ac+\dfrac bc\right)=3$

$\Leftrightarrow x\left(\dfrac1a+\dfrac1b+\dfrac1c\right)=3+\dfrac ba+\dfrac ca+\dfrac cb+\dfrac ab+\dfrac ac+\dfrac bc$

$\Leftrightarrow x\cdot\dfrac{ab+bc+ca}{abc}=\dfrac{3abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2}{abc}$

$\Leftrightarrow x(ab+bc+ca)=3abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$

$\Leftrightarrow x(ab+bc+ca)=3abc+(a+b+c)(ab+bc+ca)-3abc$

$\Leftrightarrow x(ab+bc+ca)=(a+b+c)(ab+bc+ca)$

$\Leftrightarrow x=a+b+c$

$\boxed{x=a+b+c}$

a,P=(x+a)(x+b)(x+c)

=) P= x³+(a+b+c)x²+(ab+bc+ca)x+abc

Mà a+b+c=12 , ab+bc+ca=17, abc=60

Nên P= x³+12x²+17x+60

\(A=\left(x+a\right)\left(x+b\right)\left(x+c\right)\\ =\left(x^2+ax+bx+ab\right)\left(x+c\right)\\ =x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\\ =x^3+6x^2-7x-60\)

\(B=\left(x+y+z\right)^2=\left[\left(x+y\right)+z\right]^2\\ =\left(x+y\right)^2+2\left(x+y\right)z+z^2\\ =x^2+2xy+y^2+2xz+2yz+z^2\\ =x^2+y^2+z^2+2xy+2yz+2zx\)

Bạn ơi! Ở bài 1 dòng thứ ba tại sao nó lại như vậy:)