a) x³+4x²+x-6=0

<=> x³+x²-2x+3x²+3x-6=0

<=>x(x²+x-2)+3(x²+x-2)=0

<=>(x+3)(x²+x-2)=0

<=>(x+3)(x²+2x-x-2)=0

<=>(x+3)[x(x+2)-(x+2)]=0

<=>(x+3)(x-1)(x+2)=0

=> x+3=0 hay

x-1=0 hay

x+2=0

<=> x=-3 hay x=1 hay x=-2

b)x³-3x²+4=0

\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4=0\)

\(\Leftrightarrow x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left\{\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

e:

Tham khảo: 

a: \(\Leftrightarrow x^2-2x+1+4x^2+4x+4-5x^2+5=0\)

\(\Leftrightarrow2x+10=0\)

hay x=-5

Mình cần gấp mn ơi ! Cảm ơn trc ạ

Mk xin lỗi nha, câu c sai đề

c) (x+6)⁴ + (x+8)⁴ = 272

a, \(x^4-2x^3+4x^2-3x+2=x^4-x^3+x^2-x^3+x^2-x+2x^2-2x+2\)

\(=x^2\left(x^2-x+1\right)-x\left(x^2-x+1\right)+2\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(x^2-x+2\right)\)

\(=\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\left(x^2-x+\frac{1}{4}+\frac{7}{4}\right)=\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\) (dpdcm)

b, \(x^6+x^5+x^4+x^2+x+1=x^4\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^4+1\right)=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\left(x^4+1\right)>0\) (đpcm)

ô ai cho bạn ấy sai zậy

a)(x+1)(x²+2x)=(x+1)x(x+2)=0

\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\x=0\\x+2=0=>x=-2\end{matrix}\right.\)

b)x(3x-2)-5(2-3x)=x(3x-2)+5(3x-2)=(3x-2)(x+5)=0

\(=>\left\{{}\begin{matrix}3x-2=0=>x=\dfrac{2}{3}\\x+5=0=>x=-5\end{matrix}\right.\)

c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)

=0

\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)

d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)

\(=>x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)

\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)