B1.Phân tích đa thức thành nhân tử
a.1 – 12xy + 36x^2y^2
b.(2x+1)^2+(3x-1)^2+2.(2x+1).(3x-1)
c.8x^3-27
d.125y^3+1
e.64x^3 -27y^3
f.27x^3 + y^3 / 8
g.(x-2)^3+64
h.125-(x+1)^3
B2.Phân tích đa thức thành nhân tử
a.x^3+6x^2+12x+8
b.(x^3-3x)^2+3x-1
c.1- 9x +27x^2-27x^3
B3.tìm x
a.4-25x^2=0
b.x^2-x+1/4=0
c.x^2-1/2 x + 1/16=0
các bạn giải nhanh giúp mik nhé , mik cảm ơn, mik cần gấp lắm
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!!!)
Đáp án:
Giải thích các bước giải:
$Bài1:_{}$
$a)1-12xy+36x^2y^2_{}$
⇔ $1^2-2.1.6xy+(6xy)^2_{}$
⇔ $(1-6xy)^2_{}$
$b)(2x+1)^2+(3x-1)^2+2(2x+1)(3x-1)_{}$
⇔ $(2x+1)^2+(3x-1)^2+2(3x+1)(3x-1)_{}$
⇔ $4x^2+4x+1+9x^2-6x+1+(4x+2)(3x-1)_{}$
⇔ $4x^2+4x+1+9x^2-6x+1+12x^2-4x+6x-2_{}$
⇔ $25x^2_{}$
⇔ $(5x)^2_{}$
$c)8x^3-27_{}$
⇔ $2^3x^3-3^3_{}$
⇔ $(2x)^3-3^3_{}$
⇔ $(2x-3)[ (2x)^2+2x.3+3^2]_{}$
⇔ $(2x-3)(4x^2+6x+9)_{}$
$d)125y^3+1_{}$
⇔ $5^3y^3+1^3_{}$
⇔ $(5y)^3+1^3_{}$
⇔ $(5y+1)[ (5y)^2-5y.1+1^2]_{}$
⇔ $(5y+1)(25y^2-5y+1)_{}$
$e)64x^3-27y^3_{}$
⇔ $4^3x^3-27^3_{}$
⇔ $(4x)^3-27^3_{}$
⇔ $(4x-27)[ (4x)^2+4x.27+27^2]_{}$
⇔ $(4x-27)(16x^2+108x+729)_{}$
$f)27x^3+_{}$ $\frac{y^3}{8}$
⇔ $3^3x^3+_{}$ $\frac{y^3}{2^3}$
⇔ $(3x)^3+_{}$ $(\frac{y}{2})^3$
⇔ $(3x+\frac{y}{2}).$ $[(3x)^2-3x. \frac{y}{2}+(\frac{y}{2})^2 ]$
⇔ $(3x+\frac{y}{2}).(9x^2-\frac{3xy}{2}+\frac{y^2}{4})_{}$
$g)(x-2)^2+64_{}$
⇔ $(x-2+4).[ (x-2)^2-(x-2).4+16]_{}$
⇔ $(x+2)[ x^2-4x+4-(4x-8)+16]_{}$
⇔ $(x+2)(x^2-4x+4-4x+8+16)_{}$
⇔ $(x+2)(x^2-8x+28)_{}$
$h)125-(x+1)^3_{}$
⇔ $[ 5-(x+1)].[ 25+5(x+1)+(x+1)^2]_{}$
⇔ $(5x-x-1)(25+5x+5+x^2+2x+1)_{}$
⇔ $(4-x)(x^2+7x+31)_{}$
$Bài2:_{}$
$a)x^3+6x^2+12x+8_{}$
⇔ $x^3+3.x^2.2+3.x.4+2^3_{}$
⇔ $x^3+3.x^2.2+3.x.2^2+2^3_{}$
⇔ $(x+2)^3_{}$
$b)x^3-3x^2+3x-1_{}$
⇔ $x^3-3.x^2.1+3.x.1-1^3_{}$
⇔ $x^3-3.x^2.1+3.x.1^2-1^3_{}$
⇔ $(x-1)^3_{}$
$c)1-9x+27x^2-27x^3_{}$
⇔ $1^3-3.1.3x+3.1.9x^2-3^3x^3_{}$
⇔ $1^3-3.1^2.3c+3.1.3^2x^2-(3x)^3_{}$
⇔ $1^3-3.1^2.3x+3.1.(3x)^2-(3x)^3_{}$
⇔ $(1-3x)^3_{}$
$Bài3:_{}$
$4-25x^2=0_{}$
⇔ $-25x^2=-4_{}$
⇔ $x^2=_{}$ $\frac{4}{25}$
⇔ $x=±_{}$ $\frac{2}{5}$
$Vậy_{}$ $x=±_{}$ $\frac{2}{5}$
$b)x^2-x_{}$ $\frac{1}{4}=0$
⇔ $4x^2-4x+1=0_{}$
⇔ $(2x-1)^2=0_{}$
⇔ $2x-1=0_{}$
⇔ $2x=1_{}$
⇔ $x=_{}$ $\frac{1}{2}$
$Vậy_{}$ $x=_{}$ $\frac{1}{2}$
$c)x^2_{}$ $-\frac{1}{2}x+$ $\frac{1}{16}=0$
⇔ $16x^2-8x+1=0_{}$
⇔ $(4x-1)^2=0_{}$
⇔ $4x-1=0_{}$
⇔ $4x=1_{}$
⇔ $x=_{}$ $\frac{1}{4}$
$Vậy_{}$ $x=_{}$ $\frac{1}{4}$