\(\left(x+3\right)\cdot\left(4-x\right)=7\)

\(\Rightarrow\left(x+3\right)\cdot\left(4-x\right)=1\cdot7=7\cdot1=\left(-1\right)\cdot\left(-7\right)=\left(-7\right)\cdot\left(-1\right)\)

TH1: \(\left(x+3\right)\cdot\left(4-x\right)=1\cdot7\)

\(\Rightarrow x+3=1\Rightarrow x=\left(-2\right)\)

\(\Rightarrow4-x=7\Rightarrow x=\left(-3\right)\) ₍₁₎

TH2 : \(\left(x+3\right)\cdot\left(4-x\right)=7\cdot1\)

\(\Rightarrow x+3=7\Rightarrow x=4\)

\(\Rightarrow4-x=1\Rightarrow x=3\) ₍₂₎

TH3 : \(\left(x+3\right)\cdot\left(4-x\right)=\left(-1\right)\cdot\left(-7\right)\)

\(\Rightarrow x+3=\left(-1\right)\Rightarrow x=\left(-4\right)\)

\(\Rightarrow4-x=\left(-7\right)\Rightarrow x=11\) ₍₃₎

TH4 : \(\left(x+3\right)\cdot\left(4-x\right)=\left(-7\right)\cdot\left(-1\right)\)

\(\Rightarrow x+3=\left(-7\right)\Rightarrow x=\left(-10\right)\)

\(\Rightarrow4-x=\left(-1\right)\Rightarrow x=5\) ₍₄₎

Từ (1),(2),(3),(4) => x ϵ ∅

ko có quả nào bởi vì trên cây táo làm gì có lê