a) \(^{x^3}\) - 7x+6=0

\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0

\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0

\(\Leftrightarrow\) x\(\left(x^2-1\right)\) - 6\(\left(x-1\right)\) =0

\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0

\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0

\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0

\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)

Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!

x⁴- 4x³+3x²+4x-4= 0

(x-1)(x+1)(x-2)²=0

x=1 ;x=-1;x=2

a)x^3 - 7x - 6

= x^3 + x^2 - x^2 - 6x - x - 6

= (x^3 + x^2) - (x^2 + x) - (6x + 6)
= x^2(x + 1) - x(x + 1) - 6(x + 1)

= (x + 1)(x^2 - x - 6)

= (x + 1)(x^2 - 3x + 2x - 6)
= (x + 1){(x^2 - 3x) + (2x - 6)}

= (x + 1){(x(x - 3) + 2(x - 3)}
= (x + 1)(x - 3)(x + 2)

1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)

Vậy: S={0;-7;8;-1}

2) Ta có: \(x^3-8x^2+17x-10=0\)

\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)

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

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

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

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

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

Vậy: S={2;1;5}

3) Ta có: \(2x^3-5x^2-x+6=0\)

\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)

4) Ta có: \(4x^4-4x^2-3=0\)

\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)

\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)

\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)

mà \(2x^2+1>0\forall x\in R\)

nên \(2x^2-3=0\)

\(\Leftrightarrow2x^2=3\)

\(\Leftrightarrow x^2=\frac{3}{2}\)

hay \(x=\pm\sqrt{\frac{3}{2}}\)

Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)

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

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

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

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

\(\left(2x-1\right).\left(2x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

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

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

\(x^3.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right).\left(x^3-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^2.\left(x-1\right)-4x^2+8x-4=0\)

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

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

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

\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)

\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)

\(\left(x-1\right).\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(\begin{cases}x=1\\x=2\end{cases}\)

Tham khảo nhé~

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a) \(x^4-4x^3+12x-9=0\)

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

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-3x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-3x+9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-3\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-3\right)\left(x-3\right)=0\)

\(\Leftrightarrow x-1=0\)hoặc \(x^2-3=0\)hoặc \(x-3=0\)

\(\Leftrightarrow x=1\)hoặc \(x=\pm\sqrt{3}\)hoặc \(x=3\)

Vậy tập nghiệm của phương trình là : \(S=\left\{1;\pm\sqrt{3};3\right\}\)

b) \(x^5-5x^3+4x=0\)

\(\Leftrightarrow x^5-x^3-4x^3+4x=0\)

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

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

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

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

\(\Leftrightarrow x=0\)hoặc \(x=\pm2\)hoặc \(x=\pm1\)

Vậy tập nghiệm của phương trình là : \(S=\left\{0;\pm2;\pm1\right\}\)

c) \(x^4-4x^3+3x^2+4x-4=0\)

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

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2+4=0\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2-x^2+4=0\right)\)

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

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

\(\Leftrightarrow x-1=0\)

hoặc \(x^2+x+2=\left(x+\frac{1}{2}^2\right)+\frac{7}{4}=0\left(ktm\right)\)

hoặc \(x-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{1;2\right\}\)

Bài 1:

\(x^4-4x^3+12x-9=0\)

\(\Rightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

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

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

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

\(\Rightarrow\left[x\left(x-3\right)-\left(x-3\right)\right]\left(x^2-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)

\(\Rightarrow\left[\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)

Bài 2:

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

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

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

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

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

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