Question

Giải phương trình:

\(\dfrac{\left(x-a\right)\cdot\left(x-c\right)}{\left(b-a\right)\cdot\left(b-c\right)}+\dfrac{\left(x-b\right)\cdot\left(x-c\right)}{\left(a-b\right)\cdot\left(a-c\right)}=1\)

a, b. c là hằng số và khác nhau đôi một

Answer 1

Bn xem lại đề bài đi ha!!!

Answer 2

\(\dfrac{ \left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)

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

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

\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(b-c-c+a\right)}=1\)

\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(-2c+b+a\right)}=1\)

\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-\left(a+b\right)\right)}{\left(b-a\right)\left(-2c+\left(a+b\right)\right)}=1\)

\(\Leftrightarrow\dfrac{\left(x-c\right)2x}{\left(b-a\right)\left(-2c\right)}=1\)

\(\Leftrightarrow\dfrac{2x^2-2xc}{-2cb+2ac}=1\)