Question

[zZz Cool Kid_new zZz]

Cho \(A=\frac{8-x}{x-3}\)

Tính A sao cho \(\frac{a+b}{x}=\frac{a+c}{13}\)và \(\frac{\left(a+c\right)^2}{\left(2a+b+c\right)\left(b-c\right)}=\frac{-169}{27}\)

Answer 1

[alibaba nguyễn]

\(\frac{a+b}{x}=\frac{a+c}{13}=\frac{b-c}{x-13}=\frac{2a+b+c}{x+13}\)

\(\Rightarrow\hept{\begin{cases}\frac{a+c}{b-c}=\frac{13}{x-13}\\\frac{a+c}{2a+b+c}=\frac{13}{x+13}\end{cases}}\)

\(\Rightarrow\frac{\left(a+c\right)^2}{\left(2a+b+c\right)\left(b-c\right)}=-\frac{169}{27}\)

\(\Leftrightarrow\frac{\left(a+c\right)}{\left(2a+b+c\right)}.\frac{\left(a+c\right)}{\left(b-c\right)}=-\frac{169}{27}\)

\(\Leftrightarrow\frac{13}{x-13}.\frac{13}{x+13}=-\frac{169}{27}\)

\(\Leftrightarrow\left(x-13\right)\left(x+13\right)=-27\)

\(\Leftrightarrow x^2-169=-27\)

\(\Leftrightarrow x^2=142\)

Làm nốt

Answer 2

[Nguyễn Linh Chi]

ĐK: x khác 0, x khác 13, x khác -13

Vì a+c khác 0 => a+b khác 0

\(\frac{a+b}{x}=\frac{a+c}{13}=\frac{2a+c+b}{x+13}=\frac{b-c}{x-13}\)

\(\Rightarrow\frac{\left(a+c\right)^2}{13^2}=\frac{2a+c+b}{x+13}.\frac{b-c}{x-13}\Rightarrow\frac{\left(a+c\right)^2}{\left(2a+c+b\right)\left(b-c\right)}=\frac{13^2}{\left(x+13\right)\left(x-13\right)}=\frac{169}{\left(x+13\right)\left(x-13\right)}\)

Từ đề ra 

=> (x+13)(x-13)=-27. Em làm tiếp nhé!