\(K=\left(x^4-6x^2+5\right)\left(x^4-4\right).\)

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

\(=\left(x^4-3x^2+2\right)\left(x^4-3x^2-10\right)\)

\(=\left(x^4-3x^2-4\right)^2-36\ge-36\forall x\)

\(\Rightarrow minK=-36\Leftrightarrow x^4-3x^2-4=0\Leftrightarrow x=\pm2\)

2x⁵ - 7x⁴ + 5x³ + 5x² - 7x + 2 = 0

<=> 2x⁵-4x⁴-3x⁴+6x³-x³+2x²+3x²-6x-x+2=0

<=> 2x⁴(x-2)-3x³(x-2)-x²(x-2)+3x(x-2)-(x-2)=0

<=>(x-2)(2x⁴-3x³-x²+3x-1)=0

<=>(x-2)(2x⁴-x³-2x³+x²-2x²+x+2x-1)=0

<=>(x-2)[x³(2x-1)-x²(2x-1)-x(2x-1)+2x-1]=0

<=>(x-2)(2x-1)(x³-x²-x+1)=0

<=>(x-2)(2x-1)[x²(x-1)-(x-1)]=0

<=>(x-2)(2x-1)(x-1)(x²-1)=0

<=>(x-2)(2x-1)(x-1)²(x+1)=0

=> x-2=0 => x=2

hoặc 2x-1=0=>x=1/2

hoặc x-1=0=>x=1

hoặc x+1=0=>x=-1

Vậy...

\(2x^5-7x^4+5x^3+5x^2-7x+2=0\)

\(\Leftrightarrow\left(2x^5-4x^4+2x^3\right)-\left(3x^4-6x^3+3x^2\right)-\left(3x^3-6x^2+3x\right)+\left(2x^2-4x+2\right)=0\)

\(\Leftrightarrow2x^3\left(x^2-2x+1\right)-3x^2\left(x^2-2x+1\right)-3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)

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

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

\(\Leftrightarrow\left(x-1\right)^2\left[2x^2\left(x+1\right)-5x\left(x+1\right)+2\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(2x^2-5x+2\right)=0\)

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

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

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

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

hoặc  \(x+1=0\)

hoặc \(x-2=0\)

hoặc \(2x-1=0\)

\(\Leftrightarrow\)\(x=1\)

hoặc \(x=-1\)

hoặc \(x=2\)

hoặc \(x=\frac{1}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-1;2;\frac{1}{2}\right\}\)

????????

https://mathoflife.wordpress.com/bat-phuong-trinh-h/ tham khảo nha

https://dota-2.vn/toan-10-003-bai-tap-giai-bat-phuong-trinh-he-bat-phuong-trinh

giông phương trình ý

3x(x-1)=1-x

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}}\)

Vậy...

Từ biểu thức, ta suy ra:

3x²-3x=1-x

<=>3x²-3x-1+x=0

<=>3x²-2x-1=0

<=>(3x²-3x)+(x-1)=0

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

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

<=>\(\orbr{\begin{cases}x=\frac{-1}{3}\\1\end{cases}}\)

Vậy phương trình có tập nghiệm S={-1/3;1}