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

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3-36x^2-54x-27-3\left(9x^2+6x+1\right)-2\left(x^2-4x+4\right)+x^2+x-2=0\)

\(\Leftrightarrow-48x^2-48x-28-27x^2-18x-3-2x^2+8x-8+x^2+x-2=0\)

\(\Leftrightarrow-76x^2-57x-41=0\)

\(\Leftrightarrow76x^2+57x+41=0\)

\(\text{Δ}=57^2-4\cdot76\cdot41=-9215< 0\)

Vậy: Phương trình vô nghiệm

(x-2)³+2.(1+2x)²=(1+x)³-3(x-2)²-(x-1)

<=>x³-6x²+12x-8+2.(1+4x+4x²)=1+3x²+3x+x³-3.(x²-4x+4)-x+1

<=>x³-6x²+12x-8+2+8x+8x²=1+3x²+3x+x³-3x²+12x-12-x+1

<=>x³+2x²+20x-6=x³+14x+2

<=>2x²+6x-8=0

<=>2x²-2x+8x-8=0

<=>2x.(x-1)+8.(x-1)=0

<=>2(x-1)(x+4)=0

<=>x-1=0 hoặc x+4=0

<=>x=1 hoặc x=-4

\(\Leftrightarrow x^3-6x^2+12x-8+3\left(4x^2-12x+9\right)=x^3+9x^2+27x+27-5\left(9x^2+6x+1\right)+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow-6x^2+12x-8+12x^2-36x+27=9x^2+27x+27-45x^2-30x-5+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2-24x+19=-36x^2-3x+22+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow42x^2-21x-3-x^2+4x-3=0\)

\(\Leftrightarrow41x^2-17x-6=0\)

\(\Delta=\left(-17\right)^2-4\cdot41\cdot\left(-6\right)=1273\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{17-\sqrt{1273}}{82}\\x_2=\dfrac{17+\sqrt{1273}}{82}\end{matrix}\right.\)

phân tích theo hằng đẳng thức rồi rút gọn là ra thôi bạn

máy tính tính dc vô nghiệm

Giải ra mk vs

Giải ra mk vs

​

\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)

\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)

\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)

\(\Leftrightarrow167x^2-255x-498=0\)

\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)

Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)

(x -2)\(^3\) +(3x-2)\(^2\) -5x (x+1) = (1+x)\(^3\) - 2(2x+1)\(^2\)

<=> (x\(^3\) -3.x\(^2\).2+3.x.2\(^2\) -2\(^3\)) + [(3x)\(^2\) - 2.3x.2 +2\(^2\)] - (5x.x+ 5x .1) = (1\(^3\) + 3.1\(^2\).x+ 3.1.x\(^2\) + x\(^3\) )- [2((2x)\(^2\) +2.2x.1+ 1\(^2\))]

<=> (x\(^3\) - 6x\(^2\) + 12x - 8) + (9x\(^2\) -12x+ 4)- (5x\(^2\) + 5x) = (1+3x + 3x\(^2\) + x\(^3\)) - [ 2.(4x\(^2\) + 4x +1]= (1+3x + 3x\(^2\) + x\(^3\)) - ( 8x\(^2\)+ 8x +2)

<=> x\(^3\) - 6x\(^2\) + 12x - 8 + 9x\(^2\) -12x+ 4 - 5x\(^2\) - 5x        = 1+3x + 3x\(^2\) + x\(^3\) -  8x\(^2\) -8x - 2

<=>  x\(^3\) +(- 6x\(^2\) + 9x\(^2\) - 5x\(^2\) ) +(12x- 12x - 5x) + (-8 +4) = (1-2) + ( 3x-8x) +( 3x\(^2\) -  8x\(^2\) ) + x\(^3\)

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

\(8x^3-12x^2+6x-1-3\left(1-6x+9x^2\right)=27+54x+36x^2+8x^3-2\left(x^2+3x-2x-6\right)\)\(8x^3-12x^2+6x-1-3+18x-27x^2=27+54x+36x^2+8x^3-2x^2-6x+4x+12\)\(8x^3-39x^2+24x-4=8x^3+34x^2+52x+39\)

\(8x^3-39x^2+24x-4-8x^3-34x^2-52x-39=0\)

\(-73x^2-28x-43=0\)

         Vậy đa thức vô nghiệm