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Từ điểm B, C vẽ các đường thẳng lần lượt đi qua AC và AB và cắt AC tại D, AB tại E. Sao cho BE = DC.
Xét tam giác BEC và tam giác DCB có:
BE = DC ( chứng minh trên )
ˆB=ˆC( giả thiết )
Cạnh BC chung
=> Tam giác BEC = tam giác DCB ( c.g.c )
Vậy nếu ˆB=ˆCthì AB = AC ( đpcm )
x³ -7x +6
= x³ -x²+x²-x-6x+6
= x²(x-1)+x(x-1)-6(x-1)
= (x-1)(x² +x-6)
= (x-1)(x²-2x+3x-6)
=(x-1)(x-2)(x+3)

Ta có:
\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)
\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)


a) => 4x2y2 - (4x2.2) yz + 4x2z2
=> 4x2.(y2+yz+z2 - 2)
chắc sai!! 45454655474675675685685787686845765756856876
a) \(4x^2y^2-8x^2yz+4x^2z^2\)
\(=\left(2xy\right)^2-2.2xy.2xz+\left(2xz\right)^2\)
\(=\left(2xy-2xz\right)^2\)
\(=4x^2\left(y-z\right)^2\)
b) \(x^8+x^7+x^6+x^5+x^3\)
\(=x^3\left(x^5+x^4+x^3+x^2+1\right)\)( có lẽ vậy )

a, x^2 + 2xy + y^2 - x - y - 12
= (x^2 + 2xy + y^2) - (x + y) - 16 + 4
= (x + y)^2 - 4^2 - (x + y - 4)
= (x + y - 4)(x + y + 4) - (x + y - 4)
= (x + y - 4)(x + y + 4 - 1)
= (x + y - 4)(x + y + 3)
b, x^6 + 27
= (x^2)^3 + 3^3
= (x^2 + 3)[(x^2)^2 - 3x^2 + 3^2]
= (x^2 + 3)(x^4 - 3x^2 + 9)
c, x^7 + x^5 + 1
=x^7 - x^6 + x^5 - x^3 + x^2 + x^6 - x^5 + x^4 - x^2 + x + x^5 - x^4 + x^3 - x + 1
= (x^2 + x + 1)(x^5 - x^4 + x^3 - x+1)

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

\(x^7+x^2+1\)
\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left[\left(x^3\right)^2-1^2\right]+\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[x^2\left(x^3+1\right)-x\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5+x^2-x^4-x+1\right)\)

\(x^7+x^2+1\)
\(=x^7+x^6+x^5+x^4+x^3+x^2+x+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
b) \(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^5-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
7 x - 6 x 2 - 2 = 4 x - 6 x 2 - 2 + 3 x = 4 x - 6 x 2 - 2 - 3 x = 2 x 2 - 3 x - 2 - 3 x = 2 x - 1 2 - 3 x