Question

tìm x,biết 

_a)7x^2+2x=0

_{b)x^2-8x+16=0}

 

 

Answer 1

a) 7x² + 2x = 0

=> x.(7x + 2) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\7x+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\7x=-2\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-2}{7}\end{cases}}\)

b) x² - 8x + 16 = 0

=> x² - 4x - 4x + 16 = 0

=> x.(x - 4) - 4.(x - 4) = 0

=> (x - 4)² = 0

=> x - 4 = 0

=> x = 4

Answer 2

a)x(7x+2)=0

=>x=0 hoặc 7x+2=0

                  7x=-2

                  x=-2/7

Vậy x=0 hoặc x=-2/7

b)x²-8x+16=0

x²-2*x*4+4²=0

(x-4)²=0

x-4=0

x=4

Vậy x=4