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a)\(x^2-6xy+9y^2-25z^2=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(5z\right)^2\)
\(=\left(x-3y\right)^2-\left(5z\right)^2=\left(x-3y-5z\right)\left(x-3y+5z\right)\)
b)\(xyz+x^2yz-6yz=yz\left(x^2+x-6\right)=yz\left(x^2+3x-2x-6\right)\)
\(=yz\left[x\left(x+3\right)-2\left(x+3\right)\right]=yz\left(x-2\right)\left(x+3\right)\)
a.x2-6xy+9y2-25z2
= ( x2-6xy+9y2)-25z2
= [x2-2x3y+(3y)2]-25z2
= (x-3y)2-252
= (x-3y+25)(x-3y-25)
\(x^4+6x^3+13x^2+12x+4\)
\(=x^4+x^3+5x^3+5x^2+8x^2+8x+4x+4\)
\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+8x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+5x^2+8x+4\right)\)
\(=\left(x+1\right)\left(x^3+x^2+4x^2+4x+4x+4\right)\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\right]\)
\(=\left(x+1\right)^2\left(x+2\right)^2\)
a: =(x^2-x+1)(x^2+x+1)
b: =x^2-6xy+9y^2=(x-3y)^2
c: =5x(x^2-2xy+y^2)
=5x(x-y)^2
d: =(x-3)^2
e: =(2y-z)(4x+7y)
a)HĐT:(x^2+1-x)(x^2+1+x)
b)=x^2-2.x.3y+(3y)^2
c)=5x(x^2-2xy+y^2)
=5x(x-y)^2
d)x^2-2.3.x+3^2
=(x-3)^2
e)(2y-z)+7y(2y-z)
=(2y-z)(1+7y)
a) \(2x^3+6xy-x^2z-3yz\)
= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)
=\(2x\left(x^2+3y\right)-z\left(x^2+3y\right)\)
=\(\left(x^2+2y\right)\left(2x-z\right)\)
b)\(x^2-6xy+9y^2-49\)
=\(x^2-2.x.3y+\left(3y\right)^2-7^2\)
=\(\left(x-3y\right)^2-7^2\)
=\(\left(x-3y+7\right)\left(x-3y-7\right)\)
Phân tích đa thức thành nhân tử:
\(^{x^2-6xy-25x^2+9y^2}\)
=\(\left(x^2-6xy+9\right)-25x^2\)
=\(\left(x-3\right)^2-25x^2\)
=\(\left(x-3-5x\right)\left(x-3+5x\right)\)
=\(\left(-4x-3\right)\left(6x-3\right)\)
Tìm x:
\(x^3+5x^2-6x=0\)
\(x\left(x^2+5x-6\right)=0\)
\(x\left(x^2-x+6x-6\right)=0\)
\(x\left[\left(x^2-x\right)+\left(6x-6\right)\right]=0\)
\(x\left[x\left(x-1\right)+6\left(x-1\right)\right]=0\)
\(x\left(x-1\right)\left(x+6\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\\x+6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=1\\x=-6\end{cases}}\)
Chỗ ngoặc "\(\hept{\begin{cases}\\\end{cases}}\)" cậu nên sử dụng ngoặc"\(\orbr{\begin{cases}\\\end{cases}}\)"