Question

[thaonguyen]

1) x³-x²-2=0

2) 4x²-12x+5=0

3) x⁴+6x²+8=0

4) x³-x²-21x+45=0

Answer 1

[Nguyễn Lê Phước Thịnh]

1) Sửa đề: \(x^3-x^2+2=0\)

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

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

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

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

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+1\ge1\ne0\forall x\)(2)

Từ (1) và (2) suy ra \(x+1=0\)

hay x=-1

Vậy: x=-1

2) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-2x-10x+5=0\)

\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)

3) Ta có: \(x^4+6x^2+8=0\)

\(\Leftrightarrow x^4+4x^2+2x^2+8=0\)

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

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

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+4\ge4\ne0\forall x\)(4)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge2\ne0\forall x\)(5)

Từ (3), (4) và (5) suy ra phương trình \(x^4+6x^2+8=0\) vô nghiệm

Vậy: x∈∅

4) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3+5x^2-6x^2-30x+9x+45=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

Vậy: x∈{-5;3}