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

=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0

=>x=1/2 hoặc x=-3 hoặc x=2

2)x^3 + x^2 + x + 1 = 0

=>.x^2(x+1)+(x+1)=0

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

=>x^2+1=0 hoặc x+1=0 

=>                      x =-1

3) 2x(x-3)+5(x-3) =0    

=>(2x+5)(x-3)=0

=>2x+5=0 hoặc x-3=0

=>x=-5/2 hoặc x=3

4)x(2x-7)-(4x-14)=0

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

=> x-2 =0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

5)2x^3+3x^2+2x+3=0

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

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

=>x^2+1=0 hoặc 2x+3=0

=>                      x =-3/2

x = 3/2 đó mình chắc chắn 100 %

X= 3/2 nha bạn, chắc chắn đó.

1) \(2x\left(x-3\right)+5x-15=0\)

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

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

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

2) \(x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

3) \(x^2-12x+36=0\)

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

\(x-6=0\)

\(x=6\)

4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)

\(x^3+27-x^3+x-27=0\)

\(x=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+1\right)+14=0\)

\(\Leftrightarrow x^2-4x+3-\left(x^2-x-2\right)+14=0\)

\(\Leftrightarrow x^2-4x+17-x^2+x+2=0\)

=>-3x+19=0

hay x=19/3(nhận)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(\dfrac{x-1}{x+1}-\dfrac{x-2}{x-3}+\dfrac{14}{x^2-2x-3}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}+\dfrac{14}{\left(x+1\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)-\left(x+1\right)\left(x-2\right)+14}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Rightarrow\left(x^2-4x+3\right)-\left(x^2-x-2\right)+14=0\\ \Leftrightarrow x^2-4x+3-x^2+x+2+14=0\)

\(\Leftrightarrow-3x+19=0\\ \Leftrightarrow x=\dfrac{19}{3}\left(tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{19}{3}\right\}\)

1) \(\Rightarrow x^2+4x+4-x^2+1=9\)

\(\Rightarrow4x=4\Rightarrow x=1\)

2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)

\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)

3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)

\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)