Question

tìm x 

a/ \(^{x^2}\)+16x+60=0

b/\(9x^2\)+6x+1=0

c/ x-\(2\sqrt{x}\)-3=0

Answer 1

x²+16x+60=0

<=> x²+10x+6x+60 

<=>x(x+10)+6(x+10)

<=>(x+6).(x+10)=0

=>\(\orbr{\begin{cases}x+6=0\\x+10=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-6\\x=-10\end{cases}}\)

b/9x²+6x+1=0

<=>9x²+3x+3x+1

<=>3x(3x+1)+(3x+1)

<=>(3x+1)(3x+1)=0

=> 3x+1=0=> x= \(\frac{-1}{3}\)

c/ x-\(2\sqrt{x}\)-3=0

<=>x+\(\sqrt{x}\)-3\(\sqrt{x}\)-3

<=>\(\sqrt{x}\)(\(\sqrt{x}\)+1)-3(\(\sqrt{x}+1\))

<=>\(\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)\)=0

=>\(\orbr{\begin{cases}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{cases}}\)<=>\(\orbr{\begin{cases}\sqrt{x}=-1\\\sqrt{x}=3\end{cases}}\)=>\(\orbr{\begin{cases}x\in\Phi\\x\in\left\{9;-9\right\}\end{cases}}\)