a) $\frac{3}{x^2-3x}$ và $\frac{5}{2x-6}$
Phân tích mẫu:
$x^2-3x=x(x-3)$
$2x - 6 = 2(x - 3)$
MTC: 2x(x - 3)
=> Quy đồng:
$\frac{3}{x(x-3)}=\frac{3\cdot 2}{x(x-3)\cdot 2}=\frac{\mathrm{𝟔}}{\mathrm{𝟐}𝐱(𝐱-\mathrm{𝟑})}$
$\frac{5}{2(x-3)}=\frac{5\cdot x}{2(x-3)\cdot x}=\frac{\mathrm{𝟓}𝐱}{\mathrm{𝟐}𝐱(𝐱-\mathrm{𝟑})}$

b) $\frac{3}{x^2-4}$ và $\frac{x}{x^2-4x+4}$
- Phân tích mẫu:
x² - 4 = (x - 2)(x + 2)
x² - 4x + 4 = (x - 2)²
MTC: (x + 2)(x - 2)²
=> Quy đồng:
$\frac{3}{(x-2)(x+2)}=\frac{3(x-2)}{(x-2)(x+2)(x-2)}=\frac{\mathrm{𝟑}𝐱-\mathrm{𝟔}}{(𝐱+\mathrm{𝟐}){(𝐱-\mathrm{𝟐})}^{\mathrm{𝟐}}}$
$\frac{x}{{(x-2)}^2}=\frac{x(x+2)}{{(x-2)}^2(x+2)}=\frac{{𝐱}^{\mathrm{𝟐}}+\mathrm{𝟐}𝐱}{(𝐱+\mathrm{𝟐}){(𝐱-\mathrm{𝟐})}^{\mathrm{𝟐}}}$

c) $\frac{5}{2x+8} \mathrm{và} \frac{x+2}{3x+12}$
- Phân tích mẫu:
2x + 8 = 2(x + 4)
3x + 12 = 3(x + 4)
MTC: 6(x + 4)
=> Quy đồng:
$\frac{5}{2(x+4)}=\frac{5\cdot 3}{2(x+4)\cdot 3}=\frac{\mathrm{𝟏𝟓}}{\mathrm{𝟔}(𝐱+\mathrm{𝟒})}$
$\frac{x+2}{3(x+12)}=\frac{(x+2)\cdot 2}{3(x+4)\cdot 2}=\frac{\mathrm{𝟐}𝐱+\mathrm{𝟒}}{\mathrm{𝟔}(𝐱+\mathrm{𝟒})}$

d) $\frac{1}{4x^2-8x+4}và\mathrm{ }\frac{5}{6x^2-6x}$
- Phân tích mẫu:
$4x^2-8x+4=4(x^2-2x+1)=4{(x-1)}^2$
$6x^2-6x=6x(x-1)$
MTC: Tìm BCNN của 4 và 6 là 12. Vậy MTC = 12x(x - 1)²
=> Quy đồng:
$\frac{1}{4{(x-1)}^2}=\frac{1\cdot 3x}{4{(x-1)}^2\cdot 3x}=\frac{\mathrm{𝟑}𝐱}{\mathrm{𝟏𝟐}𝐱{(𝐱-\mathrm{𝟏})}^{\mathrm{𝟐}}}$
$\frac{5}{6x(x-1)}=\frac{5\cdot 2(x-1)}{6x(x-1)\cdot 2(x-1)}=\frac{\mathrm{𝟏𝟎}𝐱-\mathrm{𝟏𝟎}}{\mathrm{𝟏𝟐}𝐱{(𝐱-\mathrm{𝟏})}^{\mathrm{𝟐}}}$

Để quy đồng mẫu thức các phân thức, chúng ta cần phân tích các mẫu thức thành nhân tử để tìm
mẫu thức chung (MTC), sau đó nhân cả tử và mẫu của mỗi phân thức với nhân tử phụ tương ứng. 

➡️ Bước 1: Quy đồng mẫu thức câu a 
Mẫu thức 1: x2−3x=x(x−3)x squared minus 3 x equals x open paren x minus 3 close paren
?2−3?=?(?−3)

Mẫu thức 2: 2x−6=2(x−3)2 x minus 6 equals 2 open paren x minus 3 close paren
2?−6=2(?−3)

MTC=2x(x−3)cap M cap T cap C equals 2 x open paren x minus 3 close paren
???=2?(?−3)

Phân thức 1: 3x(x−3)=3⋅2x(x−3)⋅2=62x(x−3)the fraction with numerator 3 and denominator x open paren x minus 3 close paren end-fraction equals the fraction with numerator 3 center dot 2 and denominator x open paren x minus 3 close paren center dot 2 end-fraction equals the fraction with numerator 6 and denominator 2 x open paren x minus 3 close paren end-fraction
3?(?−3)=3⋅2?(?−3)⋅2=62?(?−3)

Phân thức 2: 52(x−3)=5⋅x2(x−3)⋅x=5x2x(x−3)the fraction with numerator 5 and denominator 2 open paren x minus 3 close paren end-fraction equals the fraction with numerator 5 center dot x and denominator 2 open paren x minus 3 close paren center dot x end-fraction equals the fraction with numerator 5 x and denominator 2 x open paren x minus 3 close paren end-fraction
52(?−3)=5⋅?2(?−3)⋅?=5?2?(?−3)
 

➡️ Bước 2: Quy đồng mẫu thức câu b 
Mẫu thức 1: x2−4=(x−2)(x+2)x squared minus 4 equals open paren x minus 2 close paren open paren x plus 2 close paren
?2−4=(?−2)(?+2)

Mẫu thức 2: x2−4x+4=(x−2)2x squared minus 4 x plus 4 equals open paren x minus 2 close paren squared
?2−4?+4=(?−2)2

MTC=(x+2)(x−2)2cap M cap T cap C equals open paren x plus 2 close paren open paren x minus 2 close paren squared
???=(?+2)(?−2)2

Phân thức 1: 3(x−2)(x+2)=3(x−2)(x−2)2(x+2)the fraction with numerator 3 and denominator open paren x minus 2 close paren open paren x plus 2 close paren end-fraction equals the fraction with numerator 3 open paren x minus 2 close paren and denominator open paren x minus 2 close paren squared open paren x plus 2 close paren end-fraction
3(?−2)(?+2)=3(?−2)(?−2)2(?+2)

Phân thức 2: x(x−2)2=x(x+2)(x−2)2(x+2)=x2+2x(x−2)2(x+2)the fraction with numerator x and denominator open paren x minus 2 close paren squared end-fraction equals the fraction with numerator x open paren x plus 2 close paren and denominator open paren x minus 2 close paren squared open paren x plus 2 close paren end-fraction equals the fraction with numerator x squared plus 2 x and denominator open paren x minus 2 close paren squared open paren x plus 2 close paren end-fraction
?(?−2)2=?(?+2)(?−2)2(?+2)=?2+2?(?−2)2(?+2)
 

➡️ Bước 3: Quy đồng mẫu thức câu c 
Mẫu thức 1: 2x+8=2(x+4)2 x plus 8 equals 2 open paren x plus 4 close paren
2?+8=2(?+4)

Mẫu thức 2: 3x+12=3(x+4)3 x plus 12 equals 3 open paren x plus 4 close paren
3?+12=3(?+4)

MTC=6(x+4)cap M cap T cap C equals 6 open paren x plus 4 close paren
???=6(?+4)

Phân thức 1: 5x2(x+4)=5x⋅32(x+4)⋅3=15x6(x+4)the fraction with numerator 5 x and denominator 2 open paren x plus 4 close paren end-fraction equals the fraction with numerator 5 x center dot 3 and denominator 2 open paren x plus 4 close paren center dot 3 end-fraction equals the fraction with numerator 15 x and denominator 6 open paren x plus 4 close paren end-fraction
5?2(?+4)=5?⋅32(?+4)⋅3=15?6(?+4)

Phân thức 2: x+23(x+4)=(x+2)⋅23(x+4)⋅2=2x+46(x+4)the fraction with numerator x plus 2 and denominator 3 open paren x plus 4 close paren end-fraction equals the fraction with numerator open paren x plus 2 close paren center dot 2 and denominator 3 open paren x plus 4 close paren center dot 2 end-fraction equals the fraction with numerator 2 x plus 4 and denominator 6 open paren x plus 4 close paren end-fraction
?+23(?+4)=(?+2)⋅23(?+4)⋅2=2?+46(?+4)
 

➡️ Bước 4: Quy đồng mẫu thức câu d 
Mẫu thức 1: 4x2−8x+4=4(x2−2x+1)=4(x−1)24 x squared minus 8 x plus 4 equals 4 open paren x squared minus 2 x plus 1 close paren equals 4 open paren x minus 1 close paren squared
4?2−8?+4=4(?2−2?+1)=4(?−1)2

Mẫu thức 2: 6x2−6x=6x(x−1)6 x squared minus 6 x equals 6 x open paren x minus 1 close paren
6?2−6?=6?(?−1)

MTC=BCNN(4,6)⋅x⋅(x−1)2=12x(x−1)2cap M cap T cap C equals cap B cap C cap N cap N open paren 4 comma 6 close paren center dot x center dot open paren x minus 1 close paren squared equals 12 x open paren x minus 1 close paren squared
???=????(4,6)⋅?⋅(?−1)2=12?(?−1)2

Phân thức 1: 14(x−1)2=1⋅3x4(x−1)2⋅3x=3x12x(x−1)2the fraction with numerator 1 and denominator 4 open paren x minus 1 close paren squared end-fraction equals the fraction with numerator 1 center dot 3 x and denominator 4 open paren x minus 1 close paren squared center dot 3 x end-fraction equals the fraction with numerator 3 x and denominator 12 x open paren x minus 1 close paren squared end-fraction
14(?−1)2=1⋅3?4(?−1)2⋅3?=3?12?(?−1)2

Phân thức 2: 56x(x−1)=5⋅2(x−1)6x(x−1)⋅2(x−1)=10x−1012x(x−1)2the fraction with numerator 5 and denominator 6 x open paren x minus 1 close paren end-fraction equals the fraction with numerator 5 center dot 2 open paren x minus 1 close paren and denominator 6 x open paren x minus 1 close paren center dot 2 open paren x minus 1 close paren end-fraction equals the fraction with numerator 10 x minus 10 and denominator 12 x open paren x minus 1 close paren squared end-fraction
56?(?−1)=5⋅2(?−1)6?(?−1)⋅2(?−1)=10?−1012?(?−1)2
 

✅ Đáp án: 
a) 62x(x−3)the fraction with numerator 6 and denominator 2 x open paren x minus 3 close paren end-fraction
62?(?−3)
và 5x2x(x−3)the fraction with numerator 5 x and denominator 2 x open paren x minus 3 close paren end-fraction
5?2?(?−3)
 
b) 3x−6(x+2)(x−2)2the fraction with numerator 3 x minus 6 and denominator open paren x plus 2 close paren open paren x minus 2 close paren squared end-fraction
3?−6(?+2)(?−2)2
và x2+2x(x+2)(x−2)2the fraction with numerator x squared plus 2 x and denominator open paren x plus 2 close paren open paren x minus 2 close paren squared end-fraction
?2+2?(?+2)(?−2)2
 
c) 15x6(x+4)the fraction with numerator 15 x and denominator 6 open paren x plus 4 close paren end-fraction
15?6(?+4)
và 2x+46(x+4)the fraction with numerator 2 x plus 4 and denominator 6 open paren x plus 4 close paren end-fraction
2?+46(?+4)
 
d) 3x12x(x−1)2the fraction with numerator 3 x and denominator 12 x open paren x minus 1 close paren squared end-fraction
3?12?(?−1)2
và 10x−1012x(x−1)2the fraction with numerator 10 x minus 10 and denominator 12 x open paren x minus 1 close paren squared end-fraction
10?−1012?(?−1)2
 
Bạn có muốn mình giải thích kỹ hơn về cách tìm nhân tử phụ hay cách xác định BCNN cho các hệ số không? 

 
 
 
 
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a) 3x2−3x\dfrac{3}{x^2 - 3x}x2−3x3​ và 52x−6\dfrac{5}{2x - 6}2x−65​
Ta có:
x2−3x=x(x−3)x^2 - 3x = x(x - 3)x2−3x=x(x−3)
2x−6=2(x−3)2x - 6 = 2(x - 3)2x−6=2(x−3)

Mẫu thức chung là: 2x(x−3)2x(x - 3)2x(x−3)

Quy đồng:

3x(x−3)=62x(x−3)\dfrac{3}{x(x-3)} = \dfrac{6}{2x(x-3)}x(x−3)3​=2x(x−3)6​ 52(x−3)=5x2x(x−3)\dfrac{5}{2(x-3)} = \dfrac{5x}{2x(x-3)}2(x−3)5​=2x(x−3)5x​
b) 3x2−4\dfrac{3}{x^2 - 4}x2−43​ và xx2−4x+4\dfrac{x}{x^2 - 4x + 4}x2−4x+4x​
Ta có:
x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)x2−4=(x−2)(x+2)
x2−4x+4=(x−2)2x^2 - 4x + 4 = (x - 2)^2x2−4x+4=(x−2)2

Mẫu thức chung là: (x−2)2(x+2)(x - 2)^2 (x + 2)(x−2)2(x+2)

Quy đồng:

3(x−2)(x+2)=3(x−2)(x−2)2(x+2)\dfrac{3}{(x-2)(x+2)} = \dfrac{3(x-2)}{(x-2)^2(x+2)}(x−2)(x+2)3​=(x−2)2(x+2)3(x−2)​ x(x−2)2=x(x+2)(x−2)2(x+2)\dfrac{x}{(x-2)^2} = \dfrac{x(x+2)}{(x-2)^2(x+2)}(x−2)2x​=(x−2)2(x+2)x(x+2)​
c) 5x2x+8\dfrac{5x}{2x + 8}2x+85x​ và x+23x+12\dfrac{x + 2}{3x + 12}3x+12x+2​
Ta có:
2x+8=2(x+4)2x + 8 = 2(x + 4)2x+8=2(x+4)
3x+12=3(x+4)3x + 12 = 3(x + 4)3x+12=3(x+4)

Mẫu thức chung là: 6(x+4)6(x + 4)6(x+4)

Quy đồng:

5x2(x+4)=15x6(x+4)\dfrac{5x}{2(x+4)} = \dfrac{15x}{6(x+4)}2(x+4)5x​=6(x+4)15x​ x+23(x+4)=2(x+2)6(x+4)\dfrac{x+2}{3(x+4)} = \dfrac{2(x+2)}{6(x+4)}3(x+4)x+2​=6(x+4)2(x+2)​
d) 14x2−8x+4\dfrac{1}{4x^2 - 8x + 4}4x2−8x+41​ và 56x2−6x\dfrac{5}{6x^2 - 6x}6x2−6x5​
Ta có:
4x2−8x+4=4(x−1)24x^2 - 8x + 4 = 4(x - 1)^24x2−8x+4=4(x−1)2
6x2−6x=6x(x−1)6x^2 - 6x = 6x(x - 1)6x2−6x=6x(x−1)

Mẫu thức chung là: 12x(x−1)212x(x - 1)^212x(x−1)2

Quy đồng:

14(x−1)2=3x12x(x−1)2\dfrac{1}{4(x-1)^2} = \dfrac{3x}{12x(x-1)^2}4(x−1)21​=12x(x−1)23x​ 56x(x−1)=10(x−1)12x(x−1)2\dfrac{5}{6x(x-1)} = \dfrac{10(x-1)}{12x(x-1)^2}6x(x−1)5​=12x(x−1)210(x−1)​