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Phân tích đa thức thành nhân tử:
\(36-12x+x^2\)
\(=36-6x-6x+x^2\)
\(=\left(36-6x\right)-\left(6x-x^2\right)\)
\(=6\left(6-x\right)-x\left(6-x\right)\)
\(=\left(6-x\right)\left(6-x\right)=\left(6-x\right)^2\)

\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)
\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

a) \(12x^2y-18xy^2-30y^2=6y\left(2x^2-3xy-5y\right)\)
b) \(5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)
c) \(y\left(x-z\right)+7\left(z-x\right)=y\left(x-z\right)+7\left[-\left(x-z\right)\right]\)
\(=y\left(x-z\right)-7\left(x-z\right)\)
\(\left(y-7\right)\left(x-z\right)\)
d) \(36-12x+x^2=\left(6-x\right)^2\)
e) \(\left(y-4\right)^2-9\left(y+2\right)^2=-4\left(y +5\right)\left(2y+1\right)\)

mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)

a, \(x^2+10x+25=x^2+5x+5x+25\)
\(=\left(x+5\right)^2\)
b, \(x^2-12x+36=x^2-6x-6x+36\)
\(=\left(x-6\right)^2\)
c, \(9x^2+4+12x=9x^2+6x+6x+4\)
\(=3x\left(3x+2\right)+2\left(3x+2\right)=\left(3x+2\right)^2\)
d, \(x^2+49-14x=x^2-7x-7x+49\)
\(=\left(x-7\right)^2\)
e, \(9x^4+24x^2+16=9x^4+12x^2+12x^2+16\)
\(=3x^2\left(3x^2+4\right)+4\left(3x^2+4\right)=\left(3x^2+4\right)^2\)
g,\(4x^2-12xy+9y^2=4x^2-6xy-6xy+9y^2\)
\(=2x\left(2x-3y\right)-3y\left(2x-3y\right)=\left(2x-3y\right)^2\)
Chúc bạn học tốt!!!

Ta có :
\(B=x\left(x-2\right)y\left(y+6\right)+12x^2-24x+3y^2+18y+36\)
\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y+12\right)+12\)
\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)+12\)
\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+12\)
\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+12\ge2.3+12=18\)

\(A=x^2+12x+36=\left(x+6\right)^2\)
\(B=x^2+4xy+4y^2=\left(x+2y\right)^2\)
\(C=\left(3x-7\right)^2+10\left(3x-7\right)+25=\left(3x-2\right)^2\)
\(D=8x^3-12x^2+6x-1=\left(2x-1\right)^3\)
Việc còn lại bạn tự thay vào rồi tính thôi :v
\(A=x^2+12x+36\)
\(A=x^2+2.x.6+6^2\)
\(A=\left(x+6\right)^2\)
Thay x = 64 ta được
\(A=\left(64+6\right)^2\)
\(A=70^2\)
\(A=4900\)
\(B=x^2+4xy+4y^2\)
\(B=x^2+2.x.2y+\left(2y\right)^2\)
\(B=\left(x+2y\right)^2\)
Thay x = 2,8 và y = 3,6 ta được
\(B=\left(2,8+2.3,6\right)^2\)
\(B=\left(2,8+7,2\right)^2\)
\(B=10^2\)
\(B=100\)
\(C=\left(3x-7\right)^2+10\left(3x-7\right)+25\)
\(C=\left(3x-7\right)^2+2.\left(3x-7\right).5+5^2\)
\(C=\left(3x-7+5\right)^2\)
\(C=\left(3x-2\right)^2\)
Thay x = 16 ta được
\(C=\left(3.16-2\right)^2\)
\(C=\left(48-2\right)^2\)
\(C=46^2\)
\(C=2116\)
\(D=8x^3-12x^2+6x-1\)
\(D=\left(2x\right)^3-3.\left(2x\right)^2+3.\left(2x\right)-1^3\)
\(D=\left(2x-1\right)^3\)
Thay x = -1/2 ta được
\(D=\left[2.\left(-\dfrac{1}{2}\right)-1\right]^3\)
\(D=\left(-1-1\right)^3\)
\(D=\left(-2\right)^3\)
\(D=-8\)
X=6.
Ta có: \(x^2+36=12x\)
=>\(x^2-12x+36=0\)
=>\(x^2-2\cdot x\cdot6+6^2=0\)
=>\(\left(x-6\right)^2=0\)
=>x-6=0
=>x=6