Bài 1:
a: \(y^2-9-x^2+6x\)
\(=y^2-\left(x^2-6x+9\right)\)
\(=y^2-\left(x-3\right)^2\)
=(y-x+3)(y+x-3)
b: \(25-4x^2-4xy-y^2\)
\(=25-\left(4x^2+4xy+y^2\right)\)
\(=25-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
c: \(x^2-xz+4y^2-2yz+4xy\)
\(=x^2+4xy+4y^2-z\left(x+2y\right)\)
\(=\left(x+2y\right)^2-z\left(x+2y\right)\)
=(x+2y)(x+2y-z)
d: \(3x^2+6xy-48z^2+3y^2\)
\(=3\left\lbrack x^2+2xy+y^2-16z^2\right\rbrack\)
\(=3\left\lbrack\left(x+y\right)^2-\left(4z\right)^2\right\rbrack\)
=3(x+y+4z)(x+y-4z)
e: \(x^2-z^2+4y^2-4t^2-4xy+4zt\)
\(=x^2-4xy+4y^2-z^2+4zt-4t^2\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)
=(x-2y-z+2t)(x-2y+z-2t)
f: \(x^3+2x^2y+xy^2-16x\)
\(=x\left(x^2+2xy+y^2-16\right)\)
\(=x\left\lbrack\left(x+y\right)^2-16\right\rbrack\)
=x(x+y+4)(x+y-4)
Bài 2:
a: 3x(x-3)-4x+12=0
=>3x(x-3)-4(x-3)=0
=>(x-3)(3x-4)=0
=>\(\left[\begin{array}{l}x-3=0\\ 3x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac43\end{array}\right.\)
b: \(x^2-5x=0\)
=>x(x-5)=0
=>x=0 hoặc x=5
c: \(\left(3x-2\right)^2-\left(x+2\right)^2=0\)
=>(3x-2-x-2)(3x-2+x+2)=0
=>4x(2x-4)=0
=>8x(x-2)=0
=>x(x-2)=0
=>x=0 hoặc x=2
d: \(x^2-9-4\left(x+3\right)=0\)
=>(x+3)(x-3)-4(x+3)=0
=>(x+3)(x-7)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=7\end{array}\right.\)
Bài 3:
a: \(A=x^2-4z^2-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2=\left(x-y-2z\right)\left(x-y+2z\right)\)
Khi x=-16; y=-6; z=45 thì ta có:
A=(-16+6-2*45)(-16+6+2*45)
=(-10-90)(-10+90)
=80*(-100)=-8000
b: \(B=x^2-y^2+2y-1\)
\(=x^2-\left(y-1\right)^2\)
=(x-y+1)(x+y-1)
Khi x=75; y=26 thì B=(75-26+1)(75+26-1)
=50*100
=5000
c: \(C=2x+xy^2-x^2y-2y\)
=xy(y-x)+2(x-y)
=(x-y)(-xy+2)
Thay \(x=-\frac12;y=-\frac13\) vào C, ta được:
\(C=\left(-\frac12+\frac13\right)\left(-\frac{-1}{2}\cdot\frac13+2\right)=\frac{-1}{6}\left(\frac16+2\right)=-\frac16\cdot\frac{13}{6}=-\frac{13}{36}\)
