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a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
a/\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(a^3-3a^2b+3ab^2-b^3\right)\)\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^2\)
\(=6ab^2+2b^3\)(rút gọn hết)
b/\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-2xz+2xz+2xy-3xz-3yz-3xy\right).\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Hok tốt
1/
a, x2+36=12x
<=>x2-12x+36=0
<=>(x-6)2=0
<=>x-6=0
<=>x=6
b, 5x(x-3)+3-x=0
<=>5x(x-3)-(x-3)=0
<=>(5x-1)(x-3)=0
<=>\(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}}\)
2/ Sửa đề x2z2 = y2z2
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có
\(A=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+y^2z^2\right)^2\ge0\)
Đa thức trên tương đương với đa thức:
\(\left(xy\left(x+y\right)+xyz\right)+\left(yz\left(y+z\right)+xyz\right)+\left(xz\left(x+z\right)+xyz\right)\)
=\(xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left(xy+yz+xz\right)\)
xy(x + y) + yz( y + z )+ zx( z + x ) + 3xyz
=xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z)+xyz
=zy(x + y + z) + yz(x + y + z) + xz(x + y + z)
=(x + y + z)(xy + yz + zx)
chúc bn hok tốt
b \(x^8y^8+x^4y^4+1=x^8y^8+2x^4y^4+1-x^4y^4=\left(x^4y^4\right)^2+2x^4y^4+1-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2=\left(x^4y^4-x^2y^2+1\right)\left(x^4y^4+x^2y^2+1\right)\)
c \(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz=\left(x^2y+x^2z+xyz+xy^2\right)+\left(xz^2+yz^2+xyz+y^2z\right)\)
\(=x\left(xy+xz+yz+y^2\right)+z\left(xz+yz+xy+y^2\right)=\left(x+z\right)\left(xy+xz+yz+y^2\right)\)
\(=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
a \(3xyz+x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)=3xyz+xy^2+xz^2+x^2y+yz^2+x^2z+y^2z\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xy^2+xyz+y^2z\right)+\left(xyz+xz^2+yz^2\right)\)
\(=x\left(xy+xz+yz\right)+y\left(xy+xz+yz\right)+z\left(xy+xz+yz\right)=\left(x+y+z\right)\left(xy+xz+yz\right)\)
a)(a+b+c)3 - a3 - b3 - c3
= (a+b+c-a)( a2+b2+c2+2ab+2bc+2ac-a2-ab-ac+a2) - (b+c)(b2-bc+c2)
=(b+c)(a2+ab+ac+bc)
b) x3+y3+z3-3xyz
= (x+y)3-3xy(x+y) +z3-3xyz
= (x+y+z)(x2+y2+2xy-xz-yz+z2) - 3xy(x+y+z)
=(x+y+z)( x2+y2+z2-xy-yz-xz)
Biến đổi : \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\) theo công thức tổng của hai lập pương , ta được :
\(\left(y^2+z^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]\)
Thay vào \(A\),ta có : \(A=\left(y^2+z^2\right).B\).Trong đó :
\(B=\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)\right]+\left[\left(z^2-x^2\right)^2-\left(y^2+z^2\right)^2\right]\)
\(=\left[\left(x^2+y^2\right)\left(2x^2+y^2-z^2\right)\right]+\left[\left(2z^2-x^2+y^2\right)\left(-x^2-y^2\right)\right]\)
\(=\left(x^2+y^2\right)\left(3x^2-3z^2\right)\)
Vậy \(A=3\left(y^2+z^2\right)\left(x^2+y^2\right)\left(x^2-z^2\right)\).
a. x3+y3+z3-3xyz
=(x3+3x2y+3xy2+y3)+z3+(-3xyz-3x2y-3xy2)
=((x+y)3+z3)-3xy(x+y+z)
=(x+y+z)((x+y)2-z(x+y)+z2)-3xy(x+y+z)
=(x+y+z)(x2+2xy+y2-zx-zy+z2-3xy)
=(x+y+z)(x2-xy+y2+z2-zx-zy)
b. (x2-8)2+36
=x4-16x2+64+36
=x4-16x2+100
=(x4+20x2+100)-36x2
=(x2+10)2-36x2
=(x2-6x+10)(x2+6x+10)
Chúc bạn học giỏi, k cho mình nhé!!!