Question

Tìm x biết :

a.9x² - 6x -3=0

b.x³ +9x² +27x+ 19=0

Answer 1

9x²-6x-3=0

=>9x²-9x+3x-3=0

=>(x-1)(9x-3)=0

=>x-1=0 hoặc 9x+3 = 0

=> x=1 hoặc x=-1/3

b. x³+9x²+27x+19=0

   x³+x²+8x²+8x+19x+19=0

(x+1)(x²+8x+19)=0

x+1=0 => x=-1 

x²+8x+19= x²+8x+16+3=(x+4)²+3 lớn hơn hoặc bằng 3., lớn hơn 0 với moị x

Answer 2

a, \(\Rightarrow3\left(3x^2-2x-1\right)=0\)

\(\Rightarrow3x^2-2x-1=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=1\\3x-2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\3x-2=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\Rightarrow x^3+3x^2+6x^2+9x+18x+19=0\)

\(\Rightarrow x^2\left(x+3\right)+3x\left(x+3\right)+18\left(x+3\right)-2=0\)

\(\Rightarrow\left(x+3\right)\left(x^2+3x+18\right)=2\)

Mk k co thoi gian. buoc tiep theo tu lam not nhe

Answer 3

<=>\(9x^2-6x+1-4=0\)

<=>\(\left(3x-1\right)^2-4=0\)

<=>\(\left(3x-1\right)^2=4\)

<=>\(\left(3x-1\right)^2=2^2\)

<=>3x-1=2

<=>3x=3

<=>x=1

Vậy x=1