\(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2\left(x^2+3x\right)+1\)

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

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

Chúc bạn học tốt.

\(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

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

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

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x^4+6x^3+7x^2–6x+1

=x^4+(6x^3–2x^2)+(9x^2–6x+1)

= x^4+2x^2(3x–1)+(3x–1)^2

=(x^2+3x–1)^2

TƯƠNG TỰ CÁC BÀI TRONG NÂNG CAO PHÁT TRIỂN 8

\(x^4-6x^3+7x^2-6x+1\)

\(=x^4+x^2+1-6x^3+6x^2-6x\)

\(=\left(x^2+1\right)^2-x^2-6x\left(x^2-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)-6x\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^2+x+1-6x\right)\)

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

= x⁴ - x³ + x² - 5x³ + 5x² - 5x + x² - x +1  = x² ( x² - x +1 ) - 5x ( x² - x +1 ) + x² - x +1 = ( x² - x +1 ) ( x² - 5x + 1 )

mk ko biết

x⁴+6x³+7x²-6x+1

=(x⁴-2x²+1)+(6x³-6x)+9x²

=(x²-1)²+6x(x²-1)+9x²

=(x²-1).(x²-1+6x)+9x²

=(x²+3x-1)²

x⁴+6x³+7x²-6x+1

 =(x⁴-2x²+1)+(6x³-6x)+9x²

=(x²-1)²+6x(x²-1)+9x²

=(x²-1). (x²-1+6x)+9x²

=(x²+3x-1)²

A=x^4+6x^3+7x^2–6x+1=x^4+(6x^3–2x^2)+(9x^2–6x+1)
= x^4+2x^2(3x–1)+(3x–1)^2 =(x^2+3x–1)^2

chỉnh lại tí

Đặt P(x)=x⁴+6x³+7x²- 6x+1

Đặt y=x²-1

=>y²=x⁴-2x²+1

P(x)=x⁴-2x²+1+6x³-6x+9x²

  =(x²-1)²+6x(x²-1)+9x²

Q(y)=y²+6xy+9x²

=(y+3x)²

P(x)=(x²-1+3x)²

Đặt P(x)=x⁴+6x³+7x²- 6x+1

Đặt y=x²-1

=>y²=x⁴-2x²+1

P(x)=x⁴-2x²+1+6x³-6x+9x²

  =(x²-1)²+6x(x²-1)+9x²

Q(y)=y²+6xy+9x²

=(y+3x)²

P(x)=(x²-1+3x)²

TỰ LÀM NHÉ !

.c1: F(x) = x^4 + (6x^3 - 2x^2) + 9x^2 - 6x +1) 

= x^4 + 2x^2(3x-1)+(3x-1)^2

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

 C2:giả sử x khác 0, ta có:

F(x) = x^2(x^2 +6x+7-6/x+1/2^2)

= x^2 [ ( x^2 +1/2^2 +6(x-1/2)+7]

đặt x-1/x = y, suy ra : x^2 +1/x^2 = y^2 +2. do đó đa thức trở thành:

F(x;y) = x^2(y^2+2+6y+7)

= x^2(y+3)^3

=(xy+3x)^3

=[x(x-1/x)+3x]^2

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

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:  \(P\left(x\right)=x^4+6x^3+7x^2-6x+1\)

                        \(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)

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

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