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Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
`x^2-6x+7=(x^2-6x+9)-2=(x-3)^2-(sqrt2)^2=(x-3-sqrt2)(x-3+sqrt2)`
sửa đề câu a đi
\(x^2-6x-7=x^2+x-7\left(x+1\right)=\left(x-7\right)\left(x+1\right)\)
+)\(x^3+2x^2+xy^2-4x\)
\(=x^3+xy^2+2x^2-4x\)
\(=x\left(x^2+y^2\right)+x\left(2x-2\right)\)
\(=x\left(x^2+y^2+2x-2\right)\)
+) \(x^2-6x-7\)
\(=x^2-6x+9-16\)
\(=\left(x-3\right)^2-16\)
\(=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)
mk chỉnh đề
\(x^2-x-2=x^2-2x+x-2=x\left(x-2\right)+\left(x-2\right)=\left(x+1\right)\left(x-2\right)\)
\(x^2+6x+7=x^2+6x+9-2=\left(x+3\right)^2-2=\left(x+3-\sqrt{2}\right)\left(x+3+\sqrt{2}\right)\)
\(1;x^2-x-2\)
\(=x^2-2x+x-2\)
\(=x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
\(2,x^2+6x-7\)
\(=x^2-x+7x-7\)
\(=x\left(x-1\right)+7\left(x-1\right)=\left(x+7\right)\left(x-1\right)\)
= (1 - x3 ) + ( 6x - 6x2 )
= (1 - x ).(1 + x + x2) + 6x.(1 - x)
= (1 - x).(1+x+x2 + 6x)
= (1 - x).(1 + 7x +x2 )
\(=\left(x^3-6x^2+12x-8\right)+1\\ =\left(x-2\right)^3+1\\ =\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)\\ =\left(x-1\right)\left(x^2-5x+7\right)\)
x^2+6x-7=x^2-X+7x-7
=x(X-1)+7(x-1)
=(x-1)(x+7)