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bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
Ta có: \(6x^2-7x+2=6x^2-3x-4x+2\)
\(=\left(6x^2-3x\right)-\left(4x-2\right)\)
\(=3x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(3x-2\right)\)
a) x\(^2\)+8x +15
=( x\(^2\)+3x) + ( 5x +15)
= x(x+3)+ 5 (x+3)
=(x+3) (x+5)
b)x\(^2\)-4x-12
=( x\(^2\)- 6x) +( 2x -12)
=x(x-6) + 2 (x-6)
=(x - 6) (x+2)
c)9x\(^2\)-6x-24
=(9x\(^2\)-18x)+ (12x-24)
=9x(x-2) + 12 (x -2 )
=(x-2) (9x+12)
a) \(x^2+8x+15\)
\(=x^2+8x+16-1\)
\(=\left(x^2+8x+16\right)-1\)
\(=\left(x+4\right)^2-1\)
\(=\left(x+4-1\right)\left(x+4+1\right)\)
\(=\left(x+3\right)\left(x+5\right)\)
b) \(x^2-4x-12\)
\(=x^2-4x+4-16\)
\(=\left(x^2-4x+4\right)-4^2\)
\(=\left(x-2\right)^2-4^2\)
\(=\left(x-2-4\right)\left(x-2+4\right)\)
\(=\left(x-6\right)\left(x+2\right)\)
c) \(9x^2-6x-24\)
\(=9x^2-6x+1-25\)
\(=\left(9x^2-6x+1\right)-5^2\)
\(=\left(3x-1\right)^2-5^2\)
\(=\left(3x-1-5\right)\left(3x-1+5\right)\)
\(=\left(3x-6\right)\left(3x+4\right)\)
e) Ta có: \(a^3x-ab+b-x\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2x+ax+a-b\right)\)
\(x^4y^4+4=\left(x^4y^4+4x^2y^2+4\right)-4x^2y^2=\left(x^2y^2+2\right)^2-\left(2xy\right)^2=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)
\(a^2-9a^3+81a-81\)
Phân tích đa thức thành nhân tử:
\(-\left(9a^3-a^2-81a+81\right)\)