Phân tích đa thức sau thành nhân tử
a)\(x^2+4x-y^2+4\)
b)\(3x^2+6xy+3y^2-3z^2\)
c)\(x^2-2xy+y^2-z^2+2zt-t^2\)
Giải giúp mik nhé !!Cảm ơn m.n
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(x+2-y\right)\left(x+2+y\right)\)
c: \(=\left(x-y\right)^2\)
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) 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)-3z2
=3.(x+y)2-3z2
=3.[(x+y)2-z2]
=3.(x+y-x)(x+y+z)
c) x2 – 2xy + y2 – z2 + 2zt – t2.
=(x-y)2-(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)
\(a,x^2+6x+9\)
\(=x^2+3x+3x+9\)
\(=\left(x^2+3x\right)+\left(3x+9\right)\)
\(=x.\left(x+3\right)+3.\left(x+3\right)\)
\(=\left(x+3\right).\left(x+3\right)\)
\(=\left(x+3\right)^2\)
\(b,10x-25-x^2\)
\(=-\left(x^2-2.5.x+5^2\right)\)
\(=-\left(x-5\right)^2\)
\(c,x^2+4x-y^2+4\)
\(=\left(x^2+2.2.x+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right).\left(x+2+y\right)\)
\(d,3x^2+6xy+3y^2-3z^2\)
\(=3.[\left(x^2+2xy+y^2\right)-z^2]\)
\(=3.[\left(x+y\right)^2-z^2]\)
\(=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(e,x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=[\left(x-y\right)-\left(z-t\right)].[\left(x-y\right)+\left(z-t\right)]\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
a) Ta có : 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) - 3z2
= 3(x + y)2 - 3z2
= 3[(x + y)2 - z2]
= 3(x + y - z)(x + y + z)
\(x^2+4x-y^2+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x-y+2\right)\left(x+y+2\right)\)
hk tốt
^^
a)\(=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
b)\(=\left(x+9\right)^2-\left(6x\right)^2=\left(x+9-6x\right)\left(x+9+6x\right)=\left(-5x+9\right)\left(7x+9\right)\)
c)\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\\ =\left(x-y+z-t\right)\left(x-y-z+t\right)\)
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)-3z2
=3.(x+y)2-3z2
=3.[(x+y)2-z2]
=3.(x+y-x)(x+y+z)
c) x2 – 2xy + y2 – z2 + 2zt – t2.
=(x-y)2-(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; \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2=\left(x+y-2\right)\left(x-y+2\right)\)
b; \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)
c, \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t^2\right)=\left(x-y-z+t\right)\left(x+y+z-t\right)\)
a, 3( x2 +2xy + y2 - z2) = 3((x+y)2 - z2) = 3( x+y-z)(x+y+z)
b, (x-y)2 - (z-t)2 = (x-y+z-t)(x-y-z+t)
a/3x^2+6xy+3y^2-3z^2
=3(x^2+2xy+y^2-z^2)
=3[(x+y)^2-x^2)]
=3(x+y-z)(x+y+z)
b/x^2-2xy+y^2-z^2+2zt-t^2
=(x-y)^2-(z-t)^2
=(x-y+z-t)(x-y-z+t)
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) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)