Question

[VĂN LƯƠNG NGỌC DUYÊN]

1/ Thực hiệm phép tính:

a) \(\left(\dfrac{x-4}{2x-4}+\dfrac{2}{x^2-2x}\right)\div\dfrac{x-2}{x+1}\)

b) \(\dfrac{2x+3}{x^2-2x+1}\div\dfrac{6x+9}{x^2-1}+\dfrac{2}{x^2-1}\)

2/ Cho a, b, c thỏa mãn \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\). Tính giá trị của biểu thức \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)

Answer 1

[Duy Đỗ Ngọc Tuấn]

\(\left(\dfrac{x-4}{2x-4}+\dfrac{2}{x^2-2x}\right):\dfrac{x-2}{x+1}\)

\(=\left(\dfrac{x-4}{2\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}\right).\dfrac{x+1}{x-2}\)

\(=\dfrac{x\left(x-4\right)+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)

\(=\dfrac{x^2-4x+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)

\(=\dfrac{\left(x-2\right)^2\left(x+1\right)}{2x\left(x-2\right)\left(x-2\right)}\)

\(=\dfrac{x+1}{2x}\)

Answer 2

[Phùng Khánh Linh]

Mình làm nốt bài 2 nhé :

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

⇔ \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)

⇔ \(\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)

⇔ \(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c\)

⇔ \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)