Question

tìm x thỏa mãn: x^2-9x+20=0

Answer 1

x²-5x-4x+20=0

x²-4x-5x+20=0

x(x-4)-5(x-4)=0

(x-4)(x-5)=0

\(\Rightarrow\hept{\begin{cases}x-4=0\Rightarrow x=4\\x-5=0\Rightarrow x=5\end{cases}}\)

Answer 2

tớ hỏi nốt nha

tính nhanh:20²-19²+18²-17²+...+2²-1²

Answer 3

\(x^2-9x+20=0\)

\(\Rightarrow x^2-2.x.\frac{9}{2}+\frac{81}{4}-\frac{1}{4}=0\)

\(\Rightarrow x^2-2.x.\frac{9}{2}+\left(\frac{9}{2}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2=0\)

\(\Rightarrow\left(x-\frac{9}{2}-\frac{1}{2}\right)\left(x-\frac{9}{2}+\frac{1}{2}\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy \(x=5\)hoặc   \(x=4\)