Đáp án:
Giải thích các bước giải:
$Bài1:_{}$
$a)1-12xy+36x^2y^2_{}$
⇔ $1^2-2.1.6xy+(6xy)^2_{}$
⇔ $(1-6xy)^2_{}$
$b)(2x+1)^2+(3x-1)^2+2(2x+1)(3x-1)_{}$
⇔ $(2x+1)^2+(3x-1)^2+2(3x+1)(3x-1)_{}$
⇔ $4x^2+4x+1+9x^2-6x+1+(4x+2)(3x-1)_{}$
⇔ $4x^2+4x+1+9x^2-6x+1+12x^2-4x+6x-2_{}$
⇔ $25x^2_{}$
⇔ $(5x)^2_{}$
$c)8x^3-27_{}$
⇔ $2^3x^3-3^3_{}$
⇔ $(2x)^3-3^3_{}$
⇔ $(2x-3)[ (2x)^2+2x.3+3^2]_{}$
⇔ $(2x-3)(4x^2+6x+9)_{}$
$d)125y^3+1_{}$
⇔ $5^3y^3+1^3_{}$
⇔ $(5y)^3+1^3_{}$
⇔ $(5y+1)[ (5y)^2-5y.1+1^2]_{}$
⇔ $(5y+1)(25y^2-5y+1)_{}$
$e)64x^3-27y^3_{}$
⇔ $4^3x^3-27^3_{}$
⇔ $(4x)^3-27^3_{}$
⇔ $(4x-27)[ (4x)^2+4x.27+27^2]_{}$
⇔ $(4x-27)(16x^2+108x+729)_{}$
$f)27x^3+_{}$ $\frac{y^3}{8}$
⇔ $3^3x^3+_{}$ $\frac{y^3}{2^3}$
⇔ $(3x)^3+_{}$ $(\frac{y}{2})^3$
⇔ $(3x+\frac{y}{2}).$ $[(3x)^2-3x. \frac{y}{2}+(\frac{y}{2})^2 ]$
⇔ $(3x+\frac{y}{2}).(9x^2-\frac{3xy}{2}+\frac{y^2}{4})_{}$
$g)(x-2)^2+64_{}$
⇔ $(x-2+4).[ (x-2)^2-(x-2).4+16]_{}$
⇔ $(x+2)[ x^2-4x+4-(4x-8)+16]_{}$
⇔ $(x+2)(x^2-4x+4-4x+8+16)_{}$
⇔ $(x+2)(x^2-8x+28)_{}$
$h)125-(x+1)^3_{}$
⇔ $[ 5-(x+1)].[ 25+5(x+1)+(x+1)^2]_{}$
⇔ $(5x-x-1)(25+5x+5+x^2+2x+1)_{}$
⇔ $(4-x)(x^2+7x+31)_{}$
$Bài2:_{}$
$a)x^3+6x^2+12x+8_{}$
⇔ $x^3+3.x^2.2+3.x.4+2^3_{}$
⇔ $x^3+3.x^2.2+3.x.2^2+2^3_{}$
⇔ $(x+2)^3_{}$
$b)x^3-3x^2+3x-1_{}$
⇔ $x^3-3.x^2.1+3.x.1-1^3_{}$
⇔ $x^3-3.x^2.1+3.x.1^2-1^3_{}$
⇔ $(x-1)^3_{}$
$c)1-9x+27x^2-27x^3_{}$
⇔ $1^3-3.1.3x+3.1.9x^2-3^3x^3_{}$
⇔ $1^3-3.1^2.3c+3.1.3^2x^2-(3x)^3_{}$
⇔ $1^3-3.1^2.3x+3.1.(3x)^2-(3x)^3_{}$
⇔ $(1-3x)^3_{}$
$Bài3:_{}$
$4-25x^2=0_{}$
⇔ $-25x^2=-4_{}$
⇔ $x^2=_{}$ $\frac{4}{25}$
⇔ $x=±_{}$ $\frac{2}{5}$
$Vậy_{}$ $x=±_{}$ $\frac{2}{5}$
$b)x^2-x_{}$ $\frac{1}{4}=0$
⇔ $4x^2-4x+1=0_{}$
⇔ $(2x-1)^2=0_{}$
⇔ $2x-1=0_{}$
⇔ $2x=1_{}$
⇔ $x=_{}$ $\frac{1}{2}$
$Vậy_{}$ $x=_{}$ $\frac{1}{2}$
$c)x^2_{}$ $-\frac{1}{2}x+$ $\frac{1}{16}=0$
⇔ $16x^2-8x+1=0_{}$
⇔ $(4x-1)^2=0_{}$
⇔ $4x-1=0_{}$
⇔ $4x=1_{}$
⇔ $x=_{}$ $\frac{1}{4}$
$Vậy_{}$ $x=_{}$ $\frac{1}{4}$
