Question

tìm x biết

a/ x^3-x^2-x+1=0

b/(2x^3-3)^2-(4x^2-9)=0

c/x^4+2x^3-6x-9=0

d/2(x+5)-x^2-5x=0

Answer 1

\(a)\)\(x^3-x^2-x+1=0\)

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)

Vậy \(x=1\) hoặc \(x=-1\)

Chúc bạn học tốt ~ 

Answer 2

a) x³-x²-x+1 = 0 \(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Leftrightarrow x=1\)

Answer 3

\(c)\)\(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\)\(\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

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

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

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

Hoặc \(x^2+3+2x=0\)

\(\Leftrightarrow\)\(x^2=3\)

Hoặc \(x\left(x+2\right)=-3\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Hoặc \(x;\left(x-2\right)\inƯ\left(-3\right)\)

Ta có bảng : 

\(x\)	\(1\)	\(-3\)	\(-1\)	\(3\)
\(x-2\)	\(-3\)	\(1\)	\(3\)	\(-1\)
\(x\)	\(1\)	\(-3\)	\(-1\)	\(3\)
\(x\)	\(-1\)	\(3\)	\(5\)	\(1\)

Vậy \(x\in\left\{1;-1;3;-3;5\right\}\)

Chúc bạn học tốt ~