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
𝑥2−3𝑥=𝑥(𝑥−3)

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

MTC=2x(x−3)cap M cap T cap C equals 2 x open paren x minus 3 close paren
𝑀𝑇𝐶=2𝑥(𝑥−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
3𝑥(𝑥−3)=3⋅2𝑥(𝑥−3)⋅2=62𝑥(𝑥−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(𝑥−3)=5⋅𝑥2(𝑥−3)⋅𝑥=5𝑥2𝑥(𝑥−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
𝑥2−4=(𝑥−2)(𝑥+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
𝑥2−4𝑥+4=(𝑥−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
𝑀𝑇𝐶=(𝑥+2)(𝑥−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(𝑥−2)(𝑥+2)=3(𝑥−2)(𝑥−2)2(𝑥+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
𝑥(𝑥−2)2=𝑥(𝑥+2)(𝑥−2)2(𝑥+2)=𝑥2+2𝑥(𝑥−2)2(𝑥+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
2𝑥+8=2(𝑥+4)

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

MTC=6(x+4)cap M cap T cap C equals 6 open paren x plus 4 close paren
𝑀𝑇𝐶=6(𝑥+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
5𝑥2(𝑥+4)=5𝑥⋅32(𝑥+4)⋅3=15𝑥6(𝑥+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
𝑥+23(𝑥+4)=(𝑥+2)⋅23(𝑥+4)⋅2=2𝑥+46(𝑥+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
4𝑥2−8𝑥+4=4(𝑥2−2𝑥+1)=4(𝑥−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
6𝑥2−6𝑥=6𝑥(𝑥−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
𝑀𝑇𝐶=𝐵𝐶𝑁𝑁(4,6)⋅𝑥⋅(𝑥−1)2=12𝑥(𝑥−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(𝑥−1)2=1⋅3𝑥4(𝑥−1)2⋅3𝑥=3𝑥12𝑥(𝑥−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
56𝑥(𝑥−1)=5⋅2(𝑥−1)6𝑥(𝑥−1)⋅2(𝑥−1)=10𝑥−1012𝑥(𝑥−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
62𝑥(𝑥−3)
và 5x2x(x−3)the fraction with numerator 5 x and denominator 2 x open paren x minus 3 close paren end-fraction
5𝑥2𝑥(𝑥−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
3𝑥−6(𝑥+2)(𝑥−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
𝑥2+2𝑥(𝑥+2)(𝑥−2)2
 
c) 15x6(x+4)the fraction with numerator 15 x and denominator 6 open paren x plus 4 close paren end-fraction
15𝑥6(𝑥+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
2𝑥+46(𝑥+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
3𝑥12𝑥(𝑥−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
10𝑥−1012𝑥(𝑥−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)​