Question

[nguyen ton vu]

Tìm x ,biết:

a, 9x^2 -6x -3=0

b, x^3 + 9x^2 +27x +19=0

c, x(x-5) (x+5) -(x+2) (x^2 -2x +4 )=3

                     giúp mình vs nhé!

Answer 1

[Aikawa Maiya]

\(a,9x^2-6x-3=0\)

\(\Leftrightarrow9x^2-6x+1-4=0\)

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

\(\Rightarrow3x-1=\pm2\)

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

Vậy \(x=1\) hoặc \(x=\frac{-1}{3}\)

\(b,x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^3+9x^2+27x+27-8=0\)

\(\Leftrightarrow\left(x+3\right)^3=8\)

\(\Rightarrow x+3=2\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

\(c,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=\frac{-11}{25}\)

Vậy \(x=\frac{-11}{25}\)

Answer 2

[Thanh Ngân]

\(9x^2-6x-3=0\)

<=> \(\left(3x\right)^2-2.3x.1+1-4=0\)

<=> \(\left(3x-1\right)^2-2^2=0\)

<=> \(\left(3x-3\right)\left(3x+1\right)=0\)

<=> \(\hept{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

\(x^3+9x^2+27x+19\) \(=0\)

<=>\(x^3+x^2+8x^2+8x+19x+19=0\)

<=> \(x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

<=> \(\left(x^2+8x+19\right)\left(x+1\right)=0\)

mà \(x^2+8x+19>0\)

=> \(x+1=0\)

<=> \(x=-1\)

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)

<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)

<=>  \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)

<=> \(x^3-25x-x^3+2x^2+4x-8=3\)

<=> \(2x^2-21x-8=3\)

<=> \(2x^2-21x-11=0\)

<=> \(2x^2-22x+x-11=0\)

<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)

<=> \(\left(2x+1\right)\left(x-11\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)