Question

giúp mình câu này với ạ,ghi rõ luôn ạ

Answer 1

\(\Leftrightarrow x\left(7-x\right)+\left(x+2\right)\left(x-4\right)=x+4\)

\(\Leftrightarrow7x-x^2+x^2-2x-8-x-4=0\)

=>4x-12=0

hay x=3(nhận)

Answer 2

ĐKXĐ:\(x\ne\pm4\)

\(\dfrac{x\left(7-x\right)}{x^2-16}+\dfrac{2+x}{x+4}=\dfrac{1}{x-4}\\ \Leftrightarrow\dfrac{x\left(7-x\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-4\right)\left(2+x\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}=0\\ \Leftrightarrow\dfrac{7x-x^2+2x-8+x^2-4x-x-4}{\left(x-4\right)\left(x+4\right)}=0\\ \Rightarrow4x-12=0\\ \Leftrightarrow x=3\left(tm\right)\)

Answer 3

\(ĐK:x\ne\pm4\)

\(\Leftrightarrow\dfrac{x\left(7-x\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x+4-\left(2+x\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\)

\(\Leftrightarrow x\left(7-x\right)=x+4-\left(2+x\right)\left(x-4\right)\)

\(\Leftrightarrow7x-x^2=x+4-2x+8-x^2+4x\)

\(\Leftrightarrow4x-12=0\)

 

\(\Leftrightarrow x=3\left(tm\right)\)