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Bài giải:
a) x2 + 4x – y2 + 4 = (x2 + 4x + 4) - y2
= (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
b) 3x2 + 6xy + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y – z)(x + y + z)
c) x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
48. Phân tích các đa thức sau thành nhân tử:
a) x2 + 4x – y2 + 4; b) 3x2 + 6xy + 3y2 – 3z2;
c) x2 – 2xy + y2 – z2 + 2zt – t2.
Bài giải:
a) x2 + 4x – y2 + 4 = (x2 + 4x + 4) - y2
= (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
b) 3x2 + 6xy + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y – z)(x + y + z)
c) x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
a) \(A=x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+3\right)\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
\(x^2-3x+xy-3y\)
\(=x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
\(x^2-2xy+y^2-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
\(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
x2 + y2 - 3x - 3y + 2xy
= ( x2 + 2xy + y2 ) - ( 3x + 3y )
= ( x + y )2 - 3( x + y )
= ( x + y )( x + y - 3 )
b) ( x2 - 4x )2 - 2( x - 2 )2 - 7
= ( x2 - 4x )2 - 2( x2 - 4x + 4 ) - 7 (*)
Đặt t = x2 - 4x
(*) <=> t2 - 2( t + 4 ) - 7
= t2 - 2t - 8 - 7
= t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 - 4x + 3 )( x2 - 4x - 5 )
= ( x2 - x - 3x + 3 )( x2 + x - 5x - 5 )
= [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 ) ]
= ( x - 1 )( x - 3 )( x + 1 )( x - 5 )
a) Ta có: \(x^2+y^2-3x-3y+2xy\)
\(=\left[\left(x^2+y^2+2xy\right)-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(\sqrt{x+y+1}\right)^2\)
\(=\left(x+y-1+\sqrt{x+y+1}\right)\left(x+y-1-\sqrt{x+y+1}\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
a, \(9x^3y^2-15x^2y^3=3x^2y^2\cdot\left(3x-5y\right)\)
b,\(25x^2-49y^2=\left(5x\right)^2-\left(7y\right)^2\)
\(=\left(5x-7y\right)\cdot\left(5x+7y\right)\)
c,\(x^2y-xy^2-7x+7y=\left(x^2y-xy^2\right)-\left(7x-7y\right)\)
\(=xy\left(x-y\right)-7\left(x-y\right)\)
,\(=\left(x-y\right)\cdot\left(xy-7\right)\)
d, \(x^2-2xy+y^2-9z^2=\left(x^2-2xy+y^2\right)-9z^2\)
\(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y+3z\right)\cdot\left(x-y-3z\right)\)
f) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
a) \(x^2+4x-y^2+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
a,\(15x^3y^4-20x^4y^3+30x^3y^3\)
=\(5x^3y^3\left(3y-4x+6\right)\)
b,\(x^2+10xy+25y^2\)
=\(x^2+2.x.5.y+\left(5y\right)^2\)
=\(\left(x+5y\right)^2\)
c,\(x^2-2xy+y^2-9z^2\)
=\(\left(x^2-2xy+y^2\right)-\left(3z\right)^2\)
=\(\left(x-y\right)^2-\left(3z\right)^2\)
=\(\left(x-y+3z\right)\left(x-y-3z\right)\)
chúc bn hok tốt 
Lời giải:
a) \(3x+3y-x^2-2xy-y^2\)
\(=3(x+y)-(x^2+2xy+y^2)\)
\(=3(x+y)-(x+y)^2=(x+y)(3-x-y)\)
b) \(15x^2-15xy-25x+25y\)
\(=(15x^2-15xy)-(25x-25y)\)
\(=15x(x-y)-25(x-y)=(x-y)(15x-25)\)
\(=5(x-y)(3x-5)\)