\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)

 

\(x^2-x+\frac{5}{36}\)
\(=x^2-\frac{1}{6}x-\frac{5}{6}x+\frac{5}{36}\)
\(=x\left(x-\frac{1}{6}\right)-\frac{5}{6}\left(x-\frac{1}{6}\right)\)
\(\left(x-\frac{1}{6}\right)\left(x-\frac{5}{6}\right)\)

\(B=1+3x-x^2\)
\(=-x^2+3x+1\)
\(=-\left(x^2-3x-1\right)\)
\(=-\left\{\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2-1\right\}\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}-\frac{4}{4}\right]\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)
\(Có:\left(x-\frac{3}{2}\right)^2\ge0\) \(\text{với mọi x}\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le0\text{ với mọi x}\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le0+\frac{13}{4}=\frac{13}{4}\text{ }\text{ với mọi x}\)
\(\text{GTLN của biểu thức B là }\frac{13}{4}\)
khi \(x-\frac{3}{2}=0\) hay \(x=\frac{3}{2}\)
 

\(\left(2x-3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow\left[\left(2x\right)^2-2.2x.3+3^2\right]-4\left(x^2-1^2\right)=49\)
\(\Leftrightarrow4x^2-12x+9-4x^2+4=49\)
\(\Leftrightarrow-12x+13=49\)
\(\Leftrightarrow-12x=36\)
\(\Leftrightarrow x=-3\)
\(\text{Vậy }x=-3\)

\(A=\left(3x-1\right)\left(9x^2+3x+1\right)\)
\(=\left(3x\right)^3-1^3\)
\(=27x^3-1\)
\(B=\left(5x+3\right)\left(25x^2-15x+9\right)\)
\(=\left(5x\right)^3+3^3\)
\(=125x^3+27\)
 

\(B=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(=\left[\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+3-x+1}{x+3}\right)\)
\(=\left(\frac{2x^2+6x-x^2+3x-x+3-x^2-1}{\left(x+3\right)\left(x-3\right)}\right):\frac{4}{x+3}\)
\(=\frac{8x-1}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{4}\)\(=\frac{8x-1}{4\left(x-3\right)}\)


 

\(\left(a+5\right)^2=a^2+2.a.5+5^2=a^2+10a+25\)

\(\left(1-x\right)^2=1-2x+x^2\)
\(\left(3a-1\right)^2=\left(3a\right)^2-2.3a.1+1^2=9a^2-6a+1\)

\(\left(5-3b\right)^2=5^2-2.5.3b+\left(3b\right)^2=25-30b+9b^2\)
\(\left(y-3\right).\left(y+3\right)=y^2-3^2=y^2-9\)

\(\left(a-b-c\right)^2-\left(a-b+c\right)^2\)
\(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)
\(=\left(-c-c\right)\left(a+a-b-b\right)\)

\(=-2c\left(2a-2b\right)\)
\(=-4c\left(a-b\right)\)