Đáp án:

 

Giải thích các bước giải:

$Bài1:_{}$ 

$a)1-12xy+36x^2y^2_{}$

⇔ $1^2-2.1.6xy+(6xy)^2_{}$

⇔ $(1-6xy)^2_{}$

$b)(2x+1)^2+(3x-1)^2+2(2x+1)(3x-1)_{}$

⇔ $(2x+1)^2+(3x-1)^2+2(3x+1)(3x-1)_{}$

⇔ $4x^2+4x+1+9x^2-6x+1+(4x+2)(3x-1)_{}$

⇔ $4x^2+4x+1+9x^2-6x+1+12x^2-4x+6x-2_{}$ 

⇔ $25x^2_{}$ 

⇔ $(5x)^2_{}$ 

$c)8x^3-27_{}$

⇔ $2^3x^3-3^3_{}$ 

⇔ $(2x)^3-3^3_{}$ 

⇔ $(2x-3)[ (2x)^2+2x.3+3^2]_{}$ 

⇔ $(2x-3)(4x^2+6x+9)_{}$ 

$d)125y^3+1_{}$

⇔ $5^3y^3+1^3_{}$ 

⇔ $(5y)^3+1^3_{}$ 

⇔ $(5y+1)[ (5y)^2-5y.1+1^2]_{}$ 

⇔ $(5y+1)(25y^2-5y+1)_{}$ 

$e)64x^3-27y^3_{}$

⇔ $4^3x^3-27^3_{}$ 

⇔ $(4x)^3-27^3_{}$ 

⇔ $(4x-27)[ (4x)^2+4x.27+27^2]_{}$ 

⇔ $(4x-27)(16x^2+108x+729)_{}$ 

$f)27x^3+_{}$ $\frac{y^3}{8}$

⇔ $3^3x^3+_{}$ $\frac{y^3}{2^3}$ 

⇔ $(3x)^3+_{}$ $(\frac{y}{2})^3$ 

⇔ $(3x+\frac{y}{2}).$ $[(3x)^2-3x. \frac{y}{2}+(\frac{y}{2})^2 ]$

⇔ $(3x+\frac{y}{2}).(9x^2-\frac{3xy}{2}+\frac{y^2}{4})_{}$ 

$g)(x-2)^2+64_{}$

⇔ $(x-2+4).[ (x-2)^2-(x-2).4+16]_{}$ 

⇔ $(x+2)[ x^2-4x+4-(4x-8)+16]_{}$ 

⇔ $(x+2)(x^2-4x+4-4x+8+16)_{}$ 

⇔ $(x+2)(x^2-8x+28)_{}$ 

$h)125-(x+1)^3_{}$

⇔ $[ 5-(x+1)].[ 25+5(x+1)+(x+1)^2]_{}$ 

⇔ $(5x-x-1)(25+5x+5+x^2+2x+1)_{}$ 

⇔ $(4-x)(x^2+7x+31)_{}$ 

$Bài2:_{}$

$a)x^3+6x^2+12x+8_{}$

⇔ $x^3+3.x^2.2+3.x.4+2^3_{}$ 

⇔ $x^3+3.x^2.2+3.x.2^2+2^3_{}$ 

⇔ $(x+2)^3_{}$ 

$b)x^3-3x^2+3x-1_{}$

⇔ $x^3-3.x^2.1+3.x.1-1^3_{}$ 

⇔ $x^3-3.x^2.1+3.x.1^2-1^3_{}$ 

⇔ $(x-1)^3_{}$ 

$c)1-9x+27x^2-27x^3_{}$

⇔ $1^3-3.1.3x+3.1.9x^2-3^3x^3_{}$ 

⇔ $1^3-3.1^2.3c+3.1.3^2x^2-(3x)^3_{}$ 

⇔ $1^3-3.1^2.3x+3.1.(3x)^2-(3x)^3_{}$ 

⇔ $(1-3x)^3_{}$ 

$Bài3:_{}$

$4-25x^2=0_{}$

⇔ $-25x^2=-4_{}$ 

⇔ $x^2=_{}$ $\frac{4}{25}$ 

⇔ $x=±_{}$ $\frac{2}{5}$ 

$Vậy_{}$ $x=±_{}$ $\frac{2}{5}$ 

$b)x^2-x_{}$ $\frac{1}{4}=0$

⇔ $4x^2-4x+1=0_{}$ 

⇔ $(2x-1)^2=0_{}$ 

⇔ $2x-1=0_{}$ 

⇔ $2x=1_{}$ 

⇔ $x=_{}$ $\frac{1}{2}$ 

$Vậy_{}$ $x=_{}$ $\frac{1}{2}$ 

$c)x^2_{}$ $-\frac{1}{2}x+$ $\frac{1}{16}=0$

⇔ $16x^2-8x+1=0_{}$ 

⇔ $(4x-1)^2=0_{}$ 

⇔ $4x-1=0_{}$ 

⇔ $4x=1_{}$ 

⇔ $x=_{}$ $\frac{1}{4}$ 

$Vậy_{}$ $x=_{}$ $\frac{1}{4}$ 

B1.Phân tích đa thức thành nhân tử

a, 1 - 12xy + 36x²y²

= 1 - 2.6xy + (6xy)²

= (1 - 6xy)² 

b.(2x+1)² + (3x-1)² + 2.(2x+1).(3x-1)

= (2x + 1)² + 2(2x + 1)(3x - 1) + (3x - 1)²

= (2x + 1 + 3x - 1)²

= (5x)²

c.8x³ - 27

= (2x)³ - 3³

= (2x - 3)(4x² + 6x + 9)

d.125y³ + 1

= (5y)³ + 1

= (5y + 1)(25y² - 5y + 1)

e.64x³  - 27y³

= (4x)³ - (3y)³

= (4x - 3y)(16x² + 12xy + 9y²)

f.27x³ + y³ / 8

= (3x)³ + ($\frac{y}{2}$)³

= (3x + $\frac{y}{2}$)(9x² - $\frac{3xy}{2}$ + $\frac{y²}{4}$)

g.(x-2)³ + 64

= (x - 2)³ + 4³

= (x - 2 + 4)[(x - 2)² - 4.(x - 2) + 16]

= (x + 2)(x² - 4x + 4 - 4x + 8 + 16)

= (x + 2)(x² - 8x + 28)

h.125 - (x+1)³

= 5³ - (x + 1)³

= (5 - x - 1)(5² + 5(x + 1) + (x + 1)²

= (4 - x)(25 + 5x + 5 + x² + 2x + 1)

= (4 - x)(x² + 7x + 31)

B2.Phân tích đa thức thành nhân tử

a.x³ + 6x² + 12x + 8

= x³ + 3x².2 + 3.x.2² + 2³

= (x + 2)³

b.(x³ - 3x)² + 3x - 1

= $x^{6}$ - 6$x^{4}$ + 9x² + 3x - 1

= $x^{6}$ + $x^{5}$ - $x^{5}$ - $x^{4}$ - 5$x^{4}$ - 5x³ + 5x³ + 5x² + 4x² + 4x - x - 1

= $x^{5}$(x + 1) - $x^{4}$(x + 1) - 5x³(x + 1) + 5x²(x + 1) + 4x(x + 1) - (x + 1)

= (x + 1)($x^{5}$ - $x^{4}$ - 5x³ + 5x² + 4x - 1)

c. 1- 9x +27x²-27x³

= 1 - 3.3x + 3.(3x)² - (3x)³

= (1 - 3x)³

B3.tìm x

a.4 - 25x²=0

⇔ 25x² = 4

⇔ x² = $\frac{4}{25}$

⇔ x = ±$\frac{2}{5}$

b.x² - x + 1/4 = 0

⇔ x² - 2.x.$\frac{1}{2}$ + $\frac{1}{4}$ = 0

⇔ (x - $\frac{1}{2}$)² = 0

⇔ x - $\frac{1}{2}$ = 0

⇔ x = $\frac{1}{2}$

c.x² -1/2 x + 1/16=0

⇔ x² - 2.x.$\frac{1}{4}$ + $\frac{1}{16}$ = 0

⇔ (x - $\frac{1}{4}$)² = 0

⇔ x - $\frac{1}{4}$ = 0

⇔ x = $\frac{1}{4}$

(Mk hk sửa đề như bạn kia đâu nha

Cho mk ctlhn nha!!!)