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xét \(x\ne0\)ta có :
\(M=\)\(^{x^2\cdot\left(x^2+6x+7-\frac{6}{x}+\frac{1}{x^2}\right)}\)
Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2-2+\frac{1}{x^2}\Leftrightarrow t^2+2=x^2+\frac{1}{x^2}\)
Do đó \(M=x^2\cdot\left(t^2+2+6t+7\right)\Leftrightarrow x^2\cdot\left(t^2+6t+9\right)\)
\(\Leftrightarrow M=x^2\cdot\left(t+3\right)^2\)
M=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2(x^2+3x-1)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)^2\)
Ta có: \(P\left(x\right)=x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=x^4+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
\(3x^3-7x^2+17x-5=3x^3-x^2-6x^2+2x+15x-5\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(x^3-x^2-4=x^3+x^2+2x-2x^2-2x-4\)
\(=x\left(x^2+x+2\right)=2\left(x^2+x+2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
a: Ta có: \(x^2+3x+4=0\)
\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)
Do đó: Phương trình vô nghiệm
\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+6x^3+9x^2-2x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=x^4+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
a: \(x^4-6x^2+8\)
\(=x^4-4x^2-2x^2+8\)
\(=\left(x^2-2\right)\left(x^2-4\right)=\left(x^2-2\right)\left(x-2\right)\left(x+2\right)\)
b: \(x^4-5x^2-14\)
\(=x^4-7x^2+2x^2-14\)
\(=x^2\left(x^2-7\right)+2\left(x^2-7\right)=\left(x^2-7\right)\left(x^2+2\right)\)
c: \(4x^4-7x^2+3\)
\(=4x^4-4x^2-3x^2+3\)
\(=\left(x^2-1\right)\left(4x^2-3\right)=\left(x-1\right)\left(x+1\right)\left(4x^2-3\right)\)
d: \(6x^4+7x^2+2\)
\(=6x^4+3x^2+4x^2+2\)
\(=\left(2x^2+1\right)\left(3x^2+2\right)\)
e: \(x^4-8x^2+15\)
\(=x^4-5x^2-3x^2+15\)
\(=x^2\left(x^2-5\right)-3\left(x^2-5\right)=\left(x^2-5\right)\cdot\left(x^2-3\right)\)