ừm..

à mà đúng rồi ạ ko sai

em sai ạ đừng làm theo

Đặt : \(\left\{{}\begin{matrix}b-c=x\\c-a=y\\a-b=z\end{matrix}\right.\)

→ A = \(\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)\)

→ A = 1 + \(\dfrac{ay}{bx}+\dfrac{az}{cx}+\dfrac{bx}{ay}+1+\dfrac{bz}{cy}+\dfrac{cx}{az}+\dfrac{cy}{bz}+1\)

→ A = 3 + \(\left(\dfrac{ay}{bx}+\dfrac{bx}{ay}\right)+\left(\dfrac{az}{cx}+\dfrac{cx}{az}\right)+\left(\dfrac{bz}{cy}+\dfrac{cy}{bz}\right)\)

→ A = 3 + \(\left(\dfrac{a^2y^2+b^2x^2}{abxy}\right)+\left(\dfrac{a^2z^2+c^2x^2}{acxz}\right)+\left(\dfrac{b^2z^2+c^2y^2}{bcyz}\right)\)

→ A = 3 + \(\left(\dfrac{a^2y^2+2abxy+b^2x^2-2abxy}{abxy}\right)+\left(\dfrac{a^2z^2+2acxz+c^2x^2-2acxz}{acxz}\right)\)+ \(\left(\dfrac{b^2z^2+2bcyz+c^2y^2-2bcyz}{bcyz}\right)\)

➞ A = \(3+\left(\dfrac{\left(ay+bx\right)^2}{abxy}-2\right)+\left(\dfrac{\left(az+cx\right)^2}{acxz}-2\right)+\left(\dfrac{\left(bz+cy\right)^2}{bcyz}-2\right)\)

Xét \(\left(ay+bx\right)^2=\left[a\left(c-a\right)+b\left(b-c\right)\right]^2=\left(ac-a^2+b^2-bc\right)^2\)

\(=\left[\left(ca-cb\right)-\left(a^2-b^2\right)\right]^2=\left[\left(a-b\right)\left(c-a-b\right)\right]^2=\left[\left(a-b\right)\left(c-c\right)\right]^2=0\)

Tương tự : \(\left(az+cx\right)^2=0;\left(bz+cy\right)^2=0\)

➜ A = 3 + ( 0 - 2 ) + ( 0 - 2 ) + ( 0 - 2 ) = -3

Vì ΔMNP đồng dạng Δ IKH nên :

\(\dfrac{MN}{IK}=\dfrac{NP}{KH}=\dfrac{MP}{IH}\)

Áp dụng tính chất DTSBN :

\(\dfrac{MN}{IK}=\dfrac{NP}{KH}=\dfrac{MP}{IH}=\dfrac{\left(MN+NP+MP\right)}{IK+KH+IH}=\dfrac{3+4+6}{39}=\dfrac{1}{3}\)

=> \(\left\{{}\begin{matrix}IK=3MN=3.3=9cm\\KH=3.NP=3.6=18cm\\IH=3.MP=3.4=12cm\end{matrix}\right.\)