Bài 6
1/ x²+ 10x+26+y²+ 2y
= x²+ 2.5.x+ 25+ y²+2y+1
= (x+5)²+(y+1)²
2/ z²- 6z+13+ t²+ 4t
= z²- 2.3.z+ 9+ t²+ 2.2.t+ 4
= (z-3)²+(t+2)²
3/ x²- 2xy+ 2y²+ 2y+1
= x²- 2xy+ y²+ y²+ 2y+1
= (x-y)²+(y+1)²
4/ 4x²+ 2z²- 4xz- 2z+1
= 4x²- 2.2xz+ z²+ z²- 2z+1
= (2x-z)²+(z-1)²
5/ 4x²- 12x- y²+ 2y+8
= 4x²- 2.2x.3+9- (y²- 2y+1)
= (2x-3)²- (y-1)²
6/ 4x²+ 2z²- 4zx- 2x+1
= 4x²- 2.2x.z+ z²+ z²- 2z+1
= (2x-z)²+(z-1)²
7/ (x+y+4)(x+y-4)
= [(x+y)+4][(x+y)-4]
= (x+y)²-4²
8/ (x-y+6)(x+y- 6)
= - (y-6-x)(y-6+x)
= -[(y-6)-x][(y-6)+x]
= - (y-6)² +x²
= x²- (y-6)²
9/ (y+2z-3)(y-2z-3)
= [(y-3)+ 2z][(y-3)- 2z]
= ( y-3)²- 4z²
10/ (x+ 2y+3z)(2y+ 3z-x)
= [(2y+3z)+x][(2y+3z)-x]
= (2y+3z)²- x²
Bài 7
1/ 25x²- 9=0
<=> 25x²=9
<=> x²= $\frac{9}{25}$
<=> x= $\frac{3}{5}$ hoặc x= $\frac{-3}{5}$
Vậy x ∈{ $\frac{3}{5}$; $\frac{-3}{5}$ }
2/ (x-3)²- 4=0
<=> (x-3)²= 4
TH1: x-3= 2 <=> x=5
TH2: x-3= -2 <=> x= 1
Vậy x∈ {5;1}
3/ x²- 2x=24
<=> x²- 2x- 24=0
<=> x²- 6x+4x-24=0
<=> x(x-6)+ 4(x-6)=0
<=> (x-6)(x+4)=0
TH1: x-6=0 <=> x=6
TH2: x+4=0 <=> x=-4
Vậy x ∈ {6;-4}
4/ (x+4)²- (x+1)(x-1)=16
<=> x²+ 8x+ 16- x²+ 1-16=0
<=> 8x+1=0
<=> x= $\frac{-1}{8}$
Vậy x= $\frac{-1}{8}$
5/ (2x-1)²+ (x+3)²- 5(x-7)(x+7)=0
<=> 4x²- 4x+1+ x²+ 6x+9- 5x²+245=0
<=> 2x+255=0
<=> x= $\frac{-255}{2}$
vậy x= $\frac{-255}{2}$
