\(x^4-8x^3+11x^2+8x-12=0\)

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

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

\(\Leftrightarrow x=\left\{1;-1;6;2\right\}\)

\(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(4x^2-4x\right)+\left(12x-12\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

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

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

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

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

\(\Leftrightarrow\)x - 1 =0 ; x + 1 = 0 ; x - 2 =0 hoặc x - 6 = 0

\(\Leftrightarrow\)x = 1 ; x = -1 ; x = 2 ; x=6

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)  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

b: 

1: \(\Leftrightarrow2x\left(x+2\right)=0\)

=>x=0 hoặc x=-2

b)x^3 - 6x^2 +11x-6=0

<=>x^3 - x^2 - 5x^2 +5x + 6x - 6=0

<=>x^2(x - 1) - 5x(x - 1) +6(x - 1)=0

<=>(x-1).(x^2 - 5x + 6)=0

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

<=>(x - 1).[(x(x-2)-3(x-2)]=0

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

<=>x-1=0hoac x-2=0 hoac x-3=0

<=>x=1hoac x=2 hoac x=3

bạn mua cái máy tính vinacal xog giải nghiệm ra hết thui

mình cần làm cụ thể =)))