Bài 4: Quy đồng mẫu thức nhiều phân thức

Khuong Tung Anh
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1: \(\dfrac{11}{x^4y};\dfrac{3}{xy^3}\)

\(\dfrac{11}{x^4y}=\dfrac{11\cdot y^2}{x^4y^3}=\dfrac{11y^2}{x^4y^3}\)

\(\dfrac{3}{xy^3}=\dfrac{3\cdot x^3}{xy^3\cdot x^3}=\dfrac{3x^3}{x^4y^3}\)

2: \(\dfrac{2}{3x^3y^2};\dfrac{3}{4x^7y}\)

\(\dfrac{2}{3x^3y^2}=\dfrac{2\cdot4\cdot x^4}{3x^3y^2\cdot4x^4}=\dfrac{8x^4}{12x^7y^2}\)

\(\dfrac{3}{4x^7y}=\dfrac{3\cdot3\cdot y}{4x^7y\cdot3y}=\dfrac{9y}{12x^7y^2}\)

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nguyễn thị maianh
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Nguyễn Lê Phước Thịnh
9 tháng 1 2023 lúc 22:44

a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)

 

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Ywn Han
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Nguyễn Lê Phước Thịnh
22 tháng 1 2023 lúc 23:44

Câu 8:

a: \(\dfrac{5}{6}=\dfrac{5\left(x-1\right)}{6\left(x-1\right)}\)

\(\dfrac{x-2}{3\left(x-1\right)}=\dfrac{2x-4}{6\left(x-1\right)}\)

b: \(\dfrac{2}{5x+5}=\dfrac{6}{15\left(x+1\right)}\)

\(\dfrac{x+2}{3x+3}=\dfrac{x+2}{3\left(x+1\right)}=\dfrac{5x+10}{15\left(x+1\right)}\)

c: \(\dfrac{1}{2x}=\dfrac{5\left(x+1\right)}{10x\left(x+1\right)}\)

\(\dfrac{5}{10x+10}=\dfrac{5}{10\left(x+1\right)}=\dfrac{5x}{10x\left(x+1\right)}\)

\(\dfrac{x+7}{5x^2+5x}=\dfrac{x+7}{5x\left(x+1\right)}=\dfrac{2x+14}{10x\left(x+1\right)}\)

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đức ngô
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Jinni Nguyễn
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Nguyễn Lê Phước Thịnh
5 tháng 12 2022 lúc 22:06

\(\dfrac{3}{2x+6}=\dfrac{3\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)

\(\dfrac{x-2}{x^2-9}=\dfrac{2x-4}{2\left(x-3\right)\left(x+3\right)}\)

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Anh Nguyễn
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Nguyễn Lê Phước Thịnh
3 tháng 12 2022 lúc 13:06

MTC=x^3*(x+1)^2*(x-1)

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Ngọc Khánh Lan Nguyễn
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Phương Thảo?
1 tháng 12 2022 lúc 22:50

ta có : `x^2 -x-6=(x-3)(x+2)`

           `x+2=x+2`

MTC : `(x-3)(x+2)`

NTP : `1;x-3`

quy đồng :

`1/(x^2-x-6)=(1.1)/((x^2-x-6)1)=1/((x-3)(x+2))`

`(x-5)/(x+2)=((x-5)(x-3))/((x+2)(x-3))= ((x-5)(x-3))/((x-3)(x+2))`

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Nguyễn Lê Phước Thịnh
1 tháng 12 2022 lúc 22:38

\(\dfrac{1}{x^2-x-6}=\dfrac{1}{\left(x-3\right)\left(x+2\right)}\)

\(\dfrac{x-5}{x+2}=\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+2\right)}\)

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@Thv
1 tháng 12 2022 lúc 21:24

\(a,\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\left(x-2\right)}{2\left(x^2-4\right)}=\dfrac{3x^2-6x}{2\left(x^2-4\right)}\)

\(\dfrac{x+3}{x^2-4}=\dfrac{2\left(x+3\right)}{2\left(x^2-4\right)}=\dfrac{2x+6}{2\left(x^2-4\right)}\)

\(b,\dfrac{x+5}{x^2+4x+4}=\dfrac{x+5}{\left(x+2\right)^2}=\dfrac{3\left(x+5\right)}{3\left(x+2\right)^2}=\dfrac{3x+15}{3\left(x+2\right)^2}\\ \dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}=\dfrac{x^2+2x}{3\left(x+2\right)^2}\)

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Akai Haruma
1 tháng 12 2022 lúc 21:25

Lời giải:

a.

\(\frac{3x}{2x+4}=\frac{3x(x-2)}{2(x+2)(x-2)}=\frac{3x^2-6x}{2(x^2-4)}\)

\(\frac{x+3}{x^2-4}=\frac{2(x+3)}{2(x^2-4)}\)

b.

\(\frac{x+5}{x^2+4x+4}=\frac{x+5}{(x+2)^2}=\frac{3(x+5)}{3(x+2)^2}\)

\(\frac{x}{3x+6}=\frac{x}{3(x+2)}=\frac{x(x+2)}{3(x+2)^2}\)

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Shynie_dayy!?
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@Thv
1 tháng 12 2022 lúc 21:19

\(a,\dfrac{4x+13}{5x^2-35x}+\dfrac{48-x}{2x\left(7-x\right)}\\ =\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{2x\left(x-7\right)}\\ =\dfrac{2.\left(4x+13\right)+5.\left(x-48\right)}{10x\left(x-7\right)}=\dfrac{8x+26+5x-240}{10x\left(x-7\right)}\\ =\dfrac{13x-214}{10x\left(x-7\right)}\)

\(b,\dfrac{3x+1}{x^2-2x+1}+\dfrac{-1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{3x+1}{\left(x-1\right)^2}+\dfrac{-1}{x+1}+\dfrac{-x-3}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(3x+1\right)\left(x-1\right)-\left(x-1\right)^2+\left(-x-3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+x-3x-1-x^2+2x-1-x^2-3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{1}{x+1}\)

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Nguyễn Lê Phước Thịnh
1 tháng 12 2022 lúc 22:57

a: \(\dfrac{4}{x^2-4y^2}=\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{8x}{2x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{5}{2x^3-8xy^2}=\dfrac{5}{2x\left(x^2-4y^2\right)}=\dfrac{5}{2x\left(x-2y\right)\left(x+2y\right)}\)

b: \(\dfrac{2}{2x^2-6x}=\dfrac{2}{2x\left(x-3\right)}=\dfrac{1}{x\left(x-3\right)}=\dfrac{x-5}{x\left(x-3\right)\left(x-5\right)}\)

\(\dfrac{6}{x^2-8x+15}=\dfrac{6x}{x\left(x-3\right)\left(x-5\right)}\)

c: \(\dfrac{2-x}{3xy}=\dfrac{\left(2-x\right)\cdot10\cdot x^3y^2}{30x^4y^3}\)

\(\dfrac{3}{5xy^3}=\dfrac{18x^3}{30x^4y^3}\)

\(\dfrac{1-y}{10x^4}=\dfrac{\left(1-y\right)\cdot3y^3}{30x^4y^3}\)

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