Ví dụ 3: Giải phương trình : (4).

Giải: Ta có phương trình:

, phương trình này có nghiệm: .

Do vậy

,

và .

a) Ta có :\(2x^4-x^3-9x^2+13x-5=0=>\left(x-1\right)^3\left(2x+5\right)=0\)

=>\(\left\{\begin{matrix}\left(x-1\right)^3=0\\2x+5=0\end{matrix}\right.=>\left\{\begin{matrix}x-1=0\\2x=-5\end{matrix}\right.=>\left\{\begin{matrix}x=1\\x=-2,5\end{matrix}\right.\)

Vậy tập nghiệm của phương trình S={-2,5 ;1}

b)\(x^4-2x^3-11x^2+12x+36=0=>\left(x-3\right)^2\left(x+2\right)^2=0\)

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

Vậy tập nghiệm của pt là S={-2;3}

trên gg có

a) 2x - 3 > 3(x - 2)

⇔ 2x - 3 > 3x - 6

⇔ 2x - 3x > -6 + 3

⇔ -x > -3

⇔ x < 3

Vậy S = {x | x < 3}

b) (12x + 1)/12 ≤ (9x + 1)/3 - (8x + 1)/4

⇔ 12x + 1 ≤ 4(9x + 1) - 3(8x + 1)

⇔ 12x + 1 ≤ 36x + 4 - 24x - 3

⇔ 12x - 36x + 24x ≤ 4 - 3 - 1

⇔ 0x ≤ 0 (luôn đúng với mọi x)

Vậy S = R

a: =>2x-3>3x-6

=>-x>-3

=>x<3

b: =>12x+1<=36x+4-24x-3

=>12x+1<=12x+1

=>0x<=0(luôn đúng)

a) \(2x-3>3\left(x-2\right)\)

\(\Leftrightarrow2x-3>3x-6\)

\(\Leftrightarrow2x-3x>-6+3\)

\(\Leftrightarrow-x>-3\)

\(\Leftrightarrow x< 3\)

Vậy bất phương trình có nghiệm là \(x< 3\)

b) \(\dfrac{12x+1}{12}\le\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow\dfrac{12x+1}{12}\le\dfrac{\left(9x+1\right).4}{3.4}-\dfrac{\left(8x+1\right)3}{4.3}\)

\(\Leftrightarrow12x+1\le36x+4-24x-3\)

\(\Leftrightarrow12x-36x+24x\le4-3-1\)

\(\Leftrightarrow0x\le0\)

Vậy bất phương trình vô nghiệm

a) \(x^4+2x^3-12x^2-13x+42=0\)

\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)

\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

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

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

Ta có:

\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)

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

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

Vậy...........

b)\(\left(x-1\right)\left(x+2\right)\left(x+4\right)\left(x+7\right)=16\)

\(\Leftrightarrow\left(x-1\right)\left(x+7\right)\left(x+2\right)\left(x+4\right)=16\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x+8\right)-16=0\)

Đặt \(x^2+6x+0,5=t\)

\(\Rightarrow\left(t-7,5\right)\left(t+7,5\right)-16=0\)

\(\Rightarrow t^2-56,25-16=0\)

\(\Leftrightarrow t^2-72,25=0\)

\(\Leftrightarrow\left(t-\dfrac{17}{2}\right)\left(t+\dfrac{17}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{17}{2}\\t=-\dfrac{17}{2}\end{matrix}\right.\)

Xét \(t=\dfrac{17}{2}\Rightarrow x^2+6x+0,5=\dfrac{17}{2}\)

\(\Leftrightarrow x^2+6x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)

TT xét \(t=-\dfrac{17}{2}\Rightarrow x=\)-3

\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)

\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)

\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)

\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)

\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)

\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)

Câu 1:

 \(x^4+5x^3-12x^2+5x+1=x^4+7x^3+x^2-2x^3-14x^2-x+x^2+7x+1\)

\(=\left(x^4+7x^3+x^2\right)-\left(2x^3+14x^2+x\right)+\left(x^2+7x+1\right)\)

\(=x^2\left(x^2+7x+1\right)-2x\left(x^2+7x+1\right)+\left(x^2+7x+1\right)\)

\(=\left(x^2-2x+1\right)\left(x^2+7x+1\right)\)

\(=\left(x-1\right)^2\left(x^2+7x+1\right)\)

Câu 2:

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=x^4-24x^3+203x^2-720x+900-24x^2\)

\(=x^4-24x^3+179x^2-720x+900\)

\(=\left(x^4-7x^3+30x^2\right)-\left(17x^3-119x^2+510x\right)+\left(30x^2-210x+900\right)\)

\(=x^2\left(x^2-7x+30\right)-17x\left(x^2-7x+30\right)+30\left(x^2-7x+30\right)\)

\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)

\(=\left(x^2-2x-15x+30\right)\left(x^2-7x+30\right)\)

\(=\left[x\left(x-2\right)-15\left(x-2\right)\right]\left(x^2-7x+30\right)\)

\(=\left(x-15\right)\left(x-2\right)\left(x^2-7x+30\right)\)

Câu 3:

\(2x^3+11x^2+3x-36=\left(2x^3+14x^2+24x\right)-\left(3x^2+21x+36\right)\)

\(=2x\left(x^2+7x+12\right)-3\left(x^2+7x+12\right)\)

\(=\left(2x-3\right)\left(x^2+7x+12\right)\)

\(=\left(2x-3\right)\left(x^2+3x+4x+12\right)\)

\(=\left(2x-3\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

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