Violympic toán 8

Nii-chan

phân tích đa thức thành nhân tử:

g) x2+ 2xz+ 2xy + 4yz

h) ( 3x + 2)2 +( 3x+2)2 - 2(9x2 - 4)

k) x2 - m2 - 4xy + 2m + 4y2 - 1

l) bc.( a+d) .( b-c) - ac ( b+d). (a-c) + ab(c+d).(a-b)

Nguyễn Lê Phước Thịnh
6 tháng 8 2020 lúc 22:12

g) Ta có: \(x^2+2xz+2xy+4yz\)

\(=x\left(x+2z\right)+2y\left(x+2z\right)\)

\(=\left(x+2z\right)\left(x+2y\right)\)

h) Ta có: \(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)

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

\(=\left(3x+2-3x+2\right)^2=4^2=16\)

k) Ta có: \(x^2-m^2-4xy+2m+4y^2-1\)

\(=\left(x^2-4xy+4y^2\right)-\left(m^2-2m+1\right)\)

\(=\left(x-2y\right)^2-\left(m-1\right)^2\)

\(=\left(x-2y+m-1\right)\left(x-2y-m+1\right)\)

l) Ta có: \(bc\left(a+d\right)\cdot\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)

\(=bc\left(ab-ac+bd-cd\right)-ac\left(ab-bc+ad-cd\right)+ab\left(ac-bc+ad-bd\right)\)

\(=ab^2c-abc^2+b^2cd-bc^2d-a^2bc+abc^2-a^2cd+ac^2d+a^2bc-ab^2c+a^2bd-ab^2d\)

\(=b^2cd-bc^2d-a^2cd+ac^2d+a^2bd-ab^2d\)

\(=bcd\left(b-c\right)-a^2d\left(c-b\right)+ad\left(c^2-b^2\right)\)

\(=bcd\left(b-c\right)+a^2d\left(b-c\right)-ad\left(b^2-c^2\right)\)

\(=bcd\left(b-c\right)+a^2d\left(b-c\right)-ad\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left[bcd+a^2d-ad\left(b+c\right)\right]\)

\(=\left(b-c\right)\cdot\left(bcd+a^2d-abd-acd\right)\)

\(=\left(b-c\right)\left[cd\left(b-a\right)+ad\left(a-b\right)\right]\)

\(=\left(b-c\right)\left[cd\left(b-a\right)-ad\left(b-a\right)\right]\)

\(=\left(b-c\right)\left(b-a\right)\left(cd-ad\right)\)

\(=\left(b-c\right)\left(b-a\right)\cdot d\cdot\left(c-a\right)\)


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