Bài 4: Quy đồng mẫu thức nhiều phân thức

khang
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khang
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Nguyễn Hoàng Minh
13 tháng 12 2021 lúc 9:15

\(\dfrac{3x}{3x^3-27x}=\dfrac{3x}{3x\left(x^2-9\right)}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+2}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\\ \dfrac{x+1}{\left(x+2\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)

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Long
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Akai Haruma
13 tháng 12 2021 lúc 6:31

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

 

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Trần Đức Anh
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Nguyễn Hoàng Minh
12 tháng 12 2021 lúc 22:47

\(a,\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)};\dfrac{3}{x^2-9}=\dfrac{6}{2\left(x-3\right)\left(x+3\right)}\\ b,\dfrac{2x}{x^2-8x+16}=\dfrac{6x}{3\left(x-4\right)^2};\dfrac{x}{3x^2-12x}=\dfrac{1}{3x-12}=\dfrac{x-4}{3\left(x-4\right)^2}\)

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linh phạm
12 tháng 12 2021 lúc 22:50

a)\(\dfrac{5}{2x+6}=\dfrac{5}{2\left(x+3\right)}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}=\dfrac{5x-15}{2\left(x+3\right)\left(x-3\right)}\\ \dfrac{3}{x^2-9}=\dfrac{3}{\left(x-3\right)\left(x+3\right)}=\dfrac{6}{2\left(x-3\right)\left(x+3\right)}\)

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Trinhdiem
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Nguyễn Hoàng Minh
12 tháng 12 2021 lúc 9:19

Câu c mình làm rồi: Mn ơi, hướng dẫn em cách để giống mẫu đi ạ! - Hoc24

\(d,\dfrac{x}{x^3-27}=\dfrac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{x+2}{x^2-6x+9}=\dfrac{x+2}{\left(x-3\right)^2}=\dfrac{\left(x+2\right)\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{x-1}{x^2+3x+9}=\dfrac{\left(x-1\right)\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(f,\dfrac{x+2}{x^2-3x+2}=\dfrac{x+2}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(x-1\right)\left(x-2\right)\left(2x-3\right)}\\ \dfrac{x}{-2x^2+5x-3}=\dfrac{-x}{\left(2x-3\right)\left(x-1\right)}=\dfrac{-x\left(x-2\right)}{\left(2x-3\right)\left(x-1\right)\left(x-2\right)}\\ \dfrac{2x+1}{-2x^2+7x-6}=\dfrac{-\left(2x+1\right)}{\left(x-2\right)\left(2x-3\right)}=\dfrac{-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)\left(2x-3\right)}\)

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Nguyễn Việt Lâm
12 tháng 12 2021 lúc 9:19

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

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

Do đó ta quy đồng:

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

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

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Trinhdiem
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Nguyễn Hoàng Minh
12 tháng 12 2021 lúc 7:55

\(\dfrac{x+1}{2x^2-x^4}=\dfrac{x+1}{x^2\left(2-x^2\right)}=\dfrac{-\left(x+1\right)\left(x^4+2x^2+4\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\\ \dfrac{x}{x^4+2x^2+4}=\dfrac{x^3\left(x^2-2\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\\ \dfrac{2x-1}{x^7-8x}=\dfrac{2x-1}{x\left(x^6-8\right)}=\dfrac{x\left(2x-1\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)

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Trinhdiem
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Nguyễn Hoàng Minh
12 tháng 12 2021 lúc 8:06

\(a,\dfrac{a+x}{axb^3}=\dfrac{a\left(a+x\right)}{a^2xb^3};\dfrac{b+x}{a^2xb^2}=\dfrac{b\left(b+x\right)}{a^2xb^3};\dfrac{b-a}{axb^2}=\dfrac{ab\left(b-a\right)}{a^2xb^3}\\ b,\dfrac{2x+1}{x^2-4ax+4a^2}=\dfrac{x\left(2x+1\right)}{x\left(x-2a\right)^2};\dfrac{x+2a}{x^2-2ax}=\dfrac{\left(x+2a\right)\left(x-2a\right)}{x\left(x-2a\right)^2}\\ c,\dfrac{a+x}{6x^2-ax-2a^2}=\dfrac{a+x}{\left(3x-2a\right)\left(2x+a\right)}=\dfrac{\left(a+x\right)\left(x+2a\right)}{\left(3x-2a\right)\left(2x+a\right)}\\ \dfrac{a-x}{3x^2+4ax-4a^2}=\dfrac{a-x}{\left(3x-2a\right)\left(x+2a\right)}=\dfrac{\left(a-x\right)\left(2x+a\right)}{\left(3x-2a\right)\left(x+2a\right)\left(2x+a\right)}\)

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DAZZㅤPC亗
10 tháng 12 2021 lúc 9:38

Mẫu thức chung: \(2x^3t^4\)

\(\left\{{}\begin{matrix}\dfrac{1}{x^3t}=\dfrac{1.2t^3}{x^3t.2t^3}=\dfrac{2t^3}{2x^3t^4}\\\dfrac{3}{2t^4}=\dfrac{3.x^3}{2t^4.x^3}=\dfrac{3x^3}{2x^3t^4}\end{matrix}\right.\)

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31. Nguyễn Thành Tài 8/1...
8 tháng 12 2021 lúc 19:45

e cần gấp ạ:(

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Nguyễn Lê Phước Thịnh
8 tháng 12 2021 lúc 20:33

8: \(\dfrac{7}{5}=\dfrac{7\left(a-3\right)}{5\left(a-3\right)}\)

\(\dfrac{7a-31}{5a-15}=\dfrac{7a-31}{5\left(a-3\right)}\)

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Nguyễn Hoàng Minh
8 tháng 12 2021 lúc 14:29

Bài 2:

\(a,B=\dfrac{-5\left(x-3\right)+18}{x-3}=-5+\dfrac{18}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(18\right)=\left\{-2;-1;1;2;3;6;9;18\right\}\left(x\in N\right)\\ \Leftrightarrow x\in\left\{1;2;4;5;6;9;12;21\right\}\\ b,x_{min}\Leftrightarrow\left(x-3\right)_{min}\Leftrightarrow x-3=-18\Leftrightarrow x=-15\)

Bài 3:

\(a,M=\dfrac{x^2+3+x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x}=\dfrac{x\left(x+1\right)}{x\left(x+3\right)}=\dfrac{x+1}{x+3}=1-\dfrac{2}{x+3}\in N\\ \Leftrightarrow x+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-5;-4;-2;-1\right\}\)

Thay vào ta thấy \(x\in\left\{-5;-4;-1\right\}\Leftrightarrow M\in N\)

\(b,M_{max}\Leftrightarrow1-\dfrac{2}{x+3}\) đạt max \(\Leftrightarrow\dfrac{2}{x+3}\) đạt min \(\Leftrightarrow x+3\) đạt max

Mà \(x\in Z\Leftrightarrow x=-1\Leftrightarrow M_{max}=1-\dfrac{2}{-1+3}=0\)

\(c,M=??\) 

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