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

\(\Leftrightarrow3\left(x^2-x+1\right)^2-2\left(x+1\right)^2=5\left(x+1\right)\left(x^2-x+1\right)\)

Đặt \(x+1=a,x^2-x+1=b\), phương trình trở thành:

\(3b^2-2a^2=5ab\) 

\(\Leftrightarrow3b^2-5ab-2a^2=0\)

\(\Leftrightarrow\)\(\left(3b+a\right)\left(b-2a\right)=0\)

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

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

Vì \(3x^2-2x+4=\left(x-1\right)^2+2x^2+3>0\forall x\)nên:

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

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

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{13}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{13}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{13}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}}}\)

Vậy phương trình có tập nghiệm: \(S=\left\{\frac{3\pm\sqrt{13}}{2}\right\}\)

\(2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\)\(\left(x^3-1\right)\)

\(\Leftrightarrow2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x-1\right)\left(x^2+x+1\right)\)

Đặt \(x-1=a,x^2+x+1=b\), phương trình trở thành:

\(2b^2-7a^2=13ab\)\(x=4\)

\(\Leftrightarrow2b^2-13ab-7a^2=0\)

\(\Leftrightarrow\left(b-7a\right)\left(a+2b\right)=0\)

\(\Leftrightarrow\left[x^2+x+1-7\left(x-1\right)\right]\left[x-1+2\left(x^2+x+1\right)\right]=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(2x^2+3x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(2x+1\right)\left(x+1\right)=0\)

-Xét các trường hợp sau:

+Với \(x-2=0\Leftrightarrow x=2\)

+Với \(x-4=0\Leftrightarrow x=4\)

+Với \(x+1=0\Leftrightarrow x=-1\)

+Với \(2x+1=0\Leftrightarrow x=-0,5\)

Vậy phương trình có tập nghiệm: \(S=\left\{-1;-0,5;2;4\right\}\)

a)3(x^2−x+1)2−2(x+1)2=5(x^3+1)

⇔3(x^4+x^2+1−2x^3+2x^2−2x)−2(x^2+2x+1)−5(x^3+1)=0

⇔3x^4−11x^3+7x^2−10x−4=0

⇔(3x^4−2x^3+4x^2)+(−9x^3+6x^2−12x)+(−3x^2+2x−4)=0

⇔x^2(3x^2−2x+4)−3x(3x^2−2x+4)−(3x^2−2x+4)=0

⇔(3x^2−2x+4)(x^2−3x−1)=0

⇔[3x^2−2x+4=0(1)x^2−3x−1=0(2)

(1)⇔Δ1=−44<0

→ phương trình (1) vô nghiệm

(2)⇔Δ2=13

→⎡⎢ ⎢ ⎢⎣x1=3+√132x^2=3−√132

b)

2(x^2+x+1)2−7(x−1)2=13(x^3−1)

⇔2(x^4+x^2+1+2x^3+2x^2+2x)−7(x^2−2x+1)−13(x^3−1)=0

⇔2x^4−9x^3−x^2+18x+8=0

⇔(2x^4−12x^3+16x^2)+(3x^3−18x^2+24x)+(x^2−6x+8)=0

⇔2x^2(x^2−6x+8)+3x(x^2−6x+8)+(x^2−6x+8)=0

⇔(x^2−6x+8)(2x^2+3x+1)=0

⇔[x^2−6x+8=0(1)2x^3+3x+1=0(2)

(1)⇔Δ1=4→√Δ1=2(1)

→[x1=4x^2=2

(2)⇔Δ2=1

→⎡⎣x3=−12x^4=−1