Question

1.Tính:

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

2.Cho \(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=\(\frac{1}{a+b+c}\)

Chứng minh rằng: \(\frac{1}{a^3}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)=\(\frac{1}{a^3+b^3+c^3}\)

3.Tìm các giá trị nguyên của x sao cho:

\(\frac{1}{x}\)+\(\frac{1}{x+2}\)+\(\frac{x-2}{x^2_{ }+2x}\)có giá trị nguyên

Answer 1

Bài 1:

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

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

\(=\frac{-ab(a-b)-bc(b-c)-ca(c-a)}{(a-b)(b-c)(c-a)}=\frac{-(a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)}{-(a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)}=1\)

Answer 2

Bài 2:

Ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

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

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

\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\)

\(\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\Rightarrow (a+b)(b+c)(c+a)=0\)

\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)

Không mất tổng quát giả sử $a+b=0$

Khi đó:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3}+\frac{1}{(-a)^3}+\frac{1}{c^3}=\frac{1}{c^3}(1)\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{a^3+(-a)^3+c^3}=\frac{1}{c^3}(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\) (đpcm)

Answer 3

Bài 3:

Điều kiện: $x\neq 0; x\neq -2$

Ta có:
\(\frac{1}{x}+\frac{1}{x+2}+\frac{x-2}{x^2+2x}=\frac{2x+2}{x(x+2)}+\frac{x-2}{x(x+2)}=\frac{3x}{x(x+2)}=\frac{3}{x+2}\)

Để biểu thức trên nhận giá trị nguyên thì $3\vdots x+2$
$\Rightarrow x+2\in\left\{\pm 1;\pm 3\right\}$

$\Leftrightarrow x\in\left\{-3;-1;1;-5\right\}$

(đều thỏa mãn)

Vậy.........