b)     \(x^3-6x^2+11x-12=0\)

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

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

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

Ta có:    \(x^2-2x+3\)

        \(=\left(x-1\right)^2+2>0\)

\(\Rightarrow\)\(x-4=0\)

\(\Leftrightarrow\)\(x=4\)

Vậy...

a)  x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(x-2)(2x-7)=0

=>x-2=0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

b, x(x-1)+2x-2=0

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

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

=>x+2=0 hoặc x-1=0

=>x=-2 hoặc x=1

c, 2x^3+3x^2+2x+3=0

=>x²(2x+3)+2x+3=0

=>(x²+1)(2x+3)=0

=>x²+1=0 hoặc 2x+3=0

Vì x²+1>0 với mọi x ->vô nghiệm

=>2x+3=0 =>x=-3/2

d, x^3+6x^2+11x+6=0

=>x³+3x³+2x+3x²+3x³+6=0

=>x(x²+3x+2)+3(x²+3x+2)=0

=>(x²+3x+2)(x+3)=0

=>[x²+x+2x+2](x+3)=0

=>[x(x+1)+2(x+1)](x+3)=0

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

=>x+1=0 hoặc x+2=0 hoặc x+3=0

giúp mình với

a)  x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(x-2)(2x-7)=0

=>x-2=0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

b, x(x-1)+2x-2=0

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

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

=>x+2=0 hoặc x-1=0

=>x=-2 hoặc x=1

c, 2x^3+3x^2+2x+3=0

=>x²(2x+3)+2x+3=0

=>(x²+1)(2x+3)=0

=>x²+1=0 hoặc 2x+3=0

Vì x²+1>0 với mọi x ->vô nghiệm

=>2x+3=0 =>x=-3/2

d, x^3+6x^2+11x+6=0

=>x³+3x³+2x+3x²+3x³+6=0

=>x(x²+3x+2)+3(x²+3x+2)=0

=>(x²+3x+2)(x+3)=0

=>[x²+x+2x+2](x+3)=0

=>[x(x+1)+2(x+1)](x+3)=0

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

=>x+1=0 hoặc x+2=0 hoặc x+3=0

=>x=-1 hoặc x=-2 hoặc x=-3

tau cung bui ma chu mi giup tao roi cam on nhe

a) Ta có: \(x^2-2x+1=0\)

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

\(\Leftrightarrow x-1=0\)hay x=1

Vậy: S={1}

c) Ta có: \(x+x^4=0\)

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

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

mà \(x^2-x+1>0\forall x\)

nên x(x+1)=0

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

Vậy: S={0;-1}

a) \(\Leftrightarrow36+3-11x=0\)

\(\Leftrightarrow-11x=-39\)

\(\Leftrightarrow x=\frac{39}{11}\)

b) \(x^2-2x\frac{1}{4}+\frac{1}{16}-\frac{81}{16}=0\)

\(\left(x-\frac{1}{4}\right)^2=\frac{81}{16}\)

\(x-\frac{1}{4}=\frac{9}{4}\)

\(x=\frac{10}{4}=\frac{5}{2}\)

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

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

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

x = 2 hoặc x = - 2 hoặc x = 3

a) \(\frac{8}{2}\)

b) \(\frac{5}{2}\)

c) x=2 hoạc x=-2 hoặc x=3

a)1-6x²-x =0<=>-(6x²+x-1)=0<=>6x²+x-1=0

<=>(6x²+3x)-(2x+1)=0<=>3x(2x+1)-(2x+1)=0

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

=>3x-1=0 hoặc 2x+1=0=>x=\(\dfrac13\) hoặc x=-\(\dfrac12\)

Vậy S={\(\dfrac13\);-\(\dfrac12\)}

b)12x²+13x+3=0<=>12x²+9x+4x+3=0<=>(12x²+9x)+(4x+3)=0

<=>3x(4x+3)+(4x+3)=0<=>(3x+1)(4x+3)=0

=>3x+1=0 hoặc 4x+3=0 <=>x=-\(\dfrac13 \) hoặc x=-\(\dfrac34\)

Vậy S={-\(\dfrac13 \);-\(\dfrac34 \)}

c)x³-11x²+30x=0<=>x(x²-11x+30)=0<=>x[(x²-6x)-(5x-30)]=0

<=>x[x(x-6)-5(x-6)]=0<=>x(x-5)(x-6)=0

=>x=0 hoặc x-5=0 hoặc x-6=0=>x=0 hoặc x=5 hoặc x=6

Vậy S={0;5;6}

d)Ta có:(x²+x+1)(x²+x+2)-12=0

Đặt:t=x²+x+1

Khi đó:a(a+1)-12=0<=>a²+a-12=0<=>(a²+4a)-(3a+12)=0

<=>a(a+4)-3(a+4)=0<=>(a-3)(a+4)=0

hay (x²+x-2)(x²+x+5)=0

<=>(x-1)(x+2)(x²+x+5)=0(x²+x-2=(x-1)(x+2))

=>x-1=0 hoặc x+2=0(vì x²+x+5=(x+\(\dfrac12\))²+\(\dfrac{19}{4}\)>0)

=>x=1 hoặc x=-2

Vậy S={1;-2}

e)Ta có:2x²+x+6>x²+x+6=(x+\(\dfrac12\))²+\(\dfrac{23}{4}\)>0

nên PT vô nghiệm

Vậy S=\(\varnothing\)

1)2x³+3x²+2x+3=0

=> (2x³+3x²)+(2x+3)=0

=> x²(2x+3)+(2x+3)=0

=> (2x+3)(x²+1)=0

=>\(\hept{\begin{cases}2x+3=0\\x^2+1=0\end{cases}}\)=>\(\hept{\begin{cases}2x=-3\\x^2=-1\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{-3}{2}\\vo.nghiem\end{cases}}\)

Vậy x=-3/2

2)x²-3x-18=0

=> (x²+3x)-(6x+18)=0

=> x(x+3)-6(x+3)=0

=> (x+3)(x-6)=0

=> \(\hept{\begin{cases}x+3=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=-3\\x=6\end{cases}}\)

Vậy x=-3 hoặc x=6

3)Sai đề rồi bạn, 30 thành 30x mới đúng

x³-11x²+30x=0

=> x(x²-11x+30)=0

=> x[(x²-5x)-(6x-30)]=0

=> x[x(x-5)-6(x-5)]=0

=> x(x-5)(x-6)=0

=>\(\hept{\begin{cases}x=0\\x-5=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=5\\x=6\end{cases}}\)

Vậy x=0 hoặc x=5 hoặc x=6

\(x^2+x-12=0\\ \Rightarrow\left(x^2+4x\right)-\left(3x+12\right)=0\\ \Rightarrow x\left(x+4\right)-3\left(x+4\right)=0\\ \Rightarrow\left(x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

a: =>(2x-1-x-3)(2x-1+x+3)=0

=>(x-4)(3x+2)=0

=>x=-2/3 hoặc x=4

b: =>-5x^2+9x=0

=>-x(5x-9)=0

=>x=0 hoặc x=9/5

c: =>2x^2-10x-x+5=0

=>(x-5)(2x-1)=0

=>x=1/2 hoặc x=5

e: =>2x(x^2-25)=0

=>x(x-5)(x+5)=0

hay \(x\in\left\{0;5;-5\right\}\)