`19,`
`a) (x+y+4)(x+y-4)`
`=(x+y)^2-4^2`
`b) (x-y+6)(x+y-6)`
`=[x-(y-6)][x+(y-6)]`
`=x^2-(y-6)^2`
`c) (y+2z-3)(y-2z-3)`
`=[(y-3)+2z][(y-3)-2z]`
`=(y-3)^2-4z^2`
`d) (x+2y+3z)(2y+3z-x)`
`=(2y+3z)^2-x^2`
`20,`
`a) (x+1)^2-(x-1)^2-3(x+1)(x-1)`
`=x^2+2x+1-(x^2-2x+1)-3(x^2-1)`
`=x^2+2x+1-x^2+2x-1-3x^2+3`
`=-3x^2+4x+3`
`b) 5(x+2)(x-2)-1/2(6-8x)^2+17`
`=5(x^2-4)-1/2(36-96x+64x^2)+17`
`=5x^2-20-9+48x-32x^2+17`
`=-27x^2+48x-12`
`21,`
`a) A=9x^2+42x+49`
`A=(3x)^2+2.3x.7+7^2`
`A=(3x+7)^2`
Thay `x=1` vào A ta có:
`A=(3.1+7)^2=10^2=100`
Vậy `A=100` với `x=1`
`b) B=25x^2-2xy+1/25y^2`
`B=(5x)^2-2.5x. 1/5y+(1/5y)^2`
`B=(5x-1/5y)^2`
Thay `x=-1/5;y=-5` ta có:
`B=(5.(-1/5)-1/5.(-5))^2=(-1+1)^2=0`
Vậy `B=0` khi `x=-1/5;y=5`
`32,`
`a) 25x^2-9=0`
`<=> (5x-3)(5x+3)=0`
`<=> [(5x-3=0),(5x+3=0):}`
`<=>`\(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{-3}{5}\end{array} \right.\)
Vậy `S={+-5/3}`
`b) (x+4)^2-(x-1)(x+1)=16`
`<=> x^2+8x+16-x^2+1=16`
`<=> 8x+1=0`
`<=> x=-1/8`
Vậy `S={-1/8}`
`c) (2x-1)^2+(x+3)^2-5(x-7)(x+7)=0`
`<=> 4x^2-4x+1+x^2+6x+9-5(x^2-49)=0`
`<=> 5x^2+2x+10-5x^2+245=0`
`<=> 2x+255=0`
`<=> x=-255/2`
Vậy `S={-255/2}`
