Question

tìm x, bt:

a)x²-7x+10=0

b)36x²-49=0

Answer 1

a) \(\Leftrightarrow x^2-5x-2x+10=0\)

\(\Leftrightarrow x\left(x-5\right)-x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

Vậy \(x=5\)hoặc \(x=2\)

b) \(\Leftrightarrow\left(6x\right)^2-7^2=0\)

\(\Leftrightarrow\left(6x+7\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}6x=-7\\6x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-7}{6}\\x=\frac{7}{6}\end{cases}}\)

Vậy \(x=\frac{-7}{6}\)hoặc \(x=\frac{7}{6}\)

Answer 2

a, x²-7x+10=0

<=> x²-2x-5x+10=0

<=> x.(x-2)-5.(x-2)=0

<=> (x-2).(x-5)=0

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)

b, 36x²-49=0

<=> (6x)²-7²=0

<=> (6x-7).(6x+7)=0

\(\Leftrightarrow\orbr{\begin{cases}6x-7=0\\6x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{6}\\x=-\frac{7}{6}\end{cases}}\)