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

                                                \(=a^3+a^2b+ab^2-a^2b-ab^2-b^3\)

                                                  \(=a^3-b^3\)

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

                                                \(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)

                                                 \(=a^3+b^3\)

(a-b)(a^2+ab+b^2)=a^3-b^3

(a+b)(a^2-ab+b^2)=a^3+b^3

đây là một hằng đẳng thức nha bạn

=a³+b³+c³-3abc

thank

1.

Cho $a+b+c=0$.

Ta có:

$a+b+c=0\Rightarrow a=-(b+c)$

$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$

$\Rightarrow a^2-b^2-c^2=2bc$

Tương tự:

$b^2-c^2-a^2=2ca$

$c^2-a^2-b^2=2ab$

Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

$=\dfrac{a^3+b^3+c^3}{2abc}$

Mà $a+b+c=0$ nên:

$a^3+b^3+c^3=3abc$

Suy ra $A=\dfrac{3abc}{2abc}$

$A=\dfrac32$

2.

Cho $abc=2$.

Ta có:

$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$

Vì $abc=2$ nên:

$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$

Suy ra:

$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$

$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$

Quy đồng và sử dụng $abc=2$:

$A=1$

Với a + b + c = 0 , ta có :

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

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

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

\(\Leftrightarrow A=\frac{ab}{-2ab}\)\(+\frac{bc}{-2bc}\)\(+\frac{ac}{-2ac}\)

\(\Leftrightarrow A=\frac{-1}{2}\)\(+\frac{-1}{2}\)\(+\frac{-1}{2}\)

\(\Leftrightarrow A=\frac{-3}{2}\)

Cko êm quỷn vở

- Phân tích ra nhân tử :

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

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

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

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

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

Từ đây ta có \(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(\Rightarrow A=a+b+c\)

a: \(B=\left(\frac{a-b}{a^2+ab}-\frac{a}{b^2+ab}\right):\left(\frac{b^3}{a^3-ab^2}+\frac{1}{a+b}\right)\)

\(=\left(\frac{a-b}{a\left(a+b\right)}-\frac{a}{b\left(a+b\right)}\right):\left(\frac{b^3}{a\left(a^2-b^2\right)}+\frac{1}{a+b}\right)\)

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

\(=\frac{ba-b^2-a^2}{ab\left(a+b\right)}\cdot\frac{a\left(a-b\right)\left(a+b\right)}{b^3+a^2-ab}\)

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)

a: \(B=\left(\frac{a-b}{a^2+ab}-\frac{a}{b^2+ab}\right):\left(\frac{b^3}{a^3-ab^2}+\frac{1}{a+b}\right)\)

\(=\left(\frac{a-b}{a\left(a+b\right)}-\frac{a}{b\left(a+b\right)}\right):\left(\frac{b^3}{a\left(a^2-b^2\right)}+\frac{1}{a+b}\right)\)

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

\(=\frac{ba-b^2-a^2}{ab\left(a+b\right)}\cdot\frac{a\left(a-b\right)\left(a+b\right)}{b^3+a^2-ab}\)

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)

\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)

    \(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)

    \(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)

    \(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)

    \(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)

    \(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)

Đề lỗi rồi chứ mình ko rút gọn đc nữa

a: \(B=\left(\frac{a-b}{a^2+ab}-\frac{a}{b^2+ab}\right):\left(\frac{b^3}{a^3-ab^2}+\frac{1}{a+b}\right)\)

\(=\left(\frac{a-b}{a\left(a+b\right)}-\frac{a}{b\left(a+b\right)}\right):\left(\frac{b^3}{a\left(a^2-b^2\right)}+\frac{1}{a+b}\right)\)

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

\(=\frac{ba-b^2-a^2}{ab\left(a+b\right)}\cdot\frac{a\left(a-b\right)\left(a+b\right)}{b^3+a^2-ab}\)

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)