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Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
k ) \(125x^3-1\)
\(=\left(5x\right)^3-1\)
\(=\left(5x-1\right)\left[\left(5x\right)^2+5x.1+1^2\right]\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m ) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right).\left[\left(x^2\right)^2+x^2.y+y^2\right]=\left(x^2-y\right).\left(x^4+x^2y+y^2\right)\)
n ) \(a^4-2a^2+1\)
\(=\left(a^2\right)^2-2.a^2.1+1^2=\left(a^2-1\right)^2\)
i ) \(a^3+6a^2+12a+8\)
\(=\left(a+2\right)^3\)
k) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
n) \(a^4-2a^2+1=\left(a^2-1\right)^2=\left(a^2-1\right)\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
i) \(a^3+6a^2+12a+8=\left(a+2\right)^2\)
\(x^2-10x+16=x^2-8x-2x+16=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)
\(x^2-2x-15=x^2-5x+3x-15=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
\(2x^2+7x+3=2x^2+x+6x+3=x\left(2x+1\right)+3\left(2x+1\right)=\left(x+3\right)\left(2x+1\right)\)
a) \(x^2-10x+16=x^2-8x-2x+16=\left(x^2-8x\right)-\left(2x-16\right)=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)b) \(x^2-2x-15=x^2+3x-5x-15=\left(x^2+3x\right)-\left(5x+15\right)=x\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-5\right)\)c) \(2x^2+7x+3=2x^2+x+6x+3=\left(2x^2+x\right)+\left(6x+3\right)=x\left(2x+1\right)+3\left(2x+1\right)=\left(2x+1\right)\left(x+3\right)\)
\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt
a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
a) 49( y - 4 )2 - 9( y + 2 )2
= 72( y - 4 )2 - 32( y + 2 )2
= ( 7y - 28 )2 - ( 3y + 6 )2
= ( 7y - 28 - 3y - 6 )( 7y - 28 + 3y + 6 )
= ( 4y - 34 )( 10y - 22 )
= 2( 2y - 17 ).2( 5y - 11 )
= 4( 2y - 17 )( 5y - 11 )
b) 3x2 - 5x - 2 = 3x2 - 6x + x - 2 = 3x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 3x + 1 )
c) x2 - 7xy + 10y2 = x2 - 2xy - 5xy + 10y2 = x( x - 2y ) - 5y( x - 2y ) = ( x - 2y )( x - 5y )
d) 3x2 - 10xy + 3y2 = 3x2 - 9xy - xy + 3y2 = 3x( x - 3y ) - y( x - 3y ) = ( x - 3y )( 3x - y )
12x2 + 7x - 12
= 12x2 + 16x - 9x - 12
= 4x.( 3x + 4 ) - 3.( 3x + 4 )
= ( 3x + 4 ).( 4x - 3 )
\(12x^2+7x-12\)
\(=12x^2+16x-9x-12\)
\(=4x\left(3x+4\right)-3\left(3x+4\right)\)
\(=\left(4x-3\right)\left(3x+4\right)\)
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
\(a^2-9-8ab+16b^2\)
\(=a^2-8ab+16b^2-9\)
\(=\left(a-4b\right)^2-9\)
\(=\left(a-4b-3\right)\left(a-4b+3\right)\)
a2 - 9 - 8ab + 16b2
⇔ (a2 - 8ab + 16b2) - 9
⇔ (a - 4b)2 - 32
⇔ (a - 4b - 3)(a - 4b + 3)