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Big City Boy
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Thịnh Gia Vân
14 tháng 12 2020 lúc 20:49

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

 

Kim Taehyungie
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Nguyễn Lê Phước Thịnh
22 tháng 2 2021 lúc 13:13

1) Ta có: \(\dfrac{x\left|x-2\right|}{x^2-5x+6}\)

\(=\left[{}\begin{matrix}\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x< 2\right)\\\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x>2\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}\dfrac{-x}{x-3}\\\dfrac{x}{x-3}\end{matrix}\right.\)

2) Ta có: \(\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)

\(=\dfrac{\left(a^x\right)^2-\left(b^x\right)^2}{a^x-b^x}\)

\(=\dfrac{\left(a^x-b^x\right)\left(a^x+b^x\right)}{a^x-b^x}=a^x+b^x\)

Dung Vu
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Nguyễn Hoàng Minh
13 tháng 11 2021 lúc 11:39

\(M=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}:2\sqrt{\dfrac{3-x+2x}{3-x}}\left(-3\le x< 3;x\ne-1\right)\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}:2\sqrt{\dfrac{x+3}{3-x}}\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}\cdot\dfrac{3-x}{2\sqrt{\left(3-x\right)}\sqrt{\left(x+3\right)}}\)

\(M=\dfrac{x+2+x\sqrt{3-x}}{x+\left(x+2\right)\sqrt{3-x}}\cdot\dfrac{\sqrt{3-x}}{2\sqrt{3-x}}\\ M=\dfrac{\left(x+2\right)\sqrt{3-x}+x\left(3-x\right)}{2x\sqrt{3-x}+2\left(x+2\right)\sqrt{3-x}}\\ M=\dfrac{\sqrt{3-x}\left(2x+2\right)}{\sqrt{3-x}\left(2x+2x+4\right)}=\dfrac{2\left(x+1\right)}{4\left(x+1\right)}=\dfrac{1}{2}\)

Dung Vu
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Nguyễn Hoàng Minh
22 tháng 11 2021 lúc 7:53

\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)

Hùng Chu
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🙂T😃r😄a😆n😂g🤣
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Nguyễn Huy Tú
18 tháng 4 2021 lúc 16:30

a, ĐK : \(x\ne1;2;3;4;5\)

b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)

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

Nguyen Hoang Khanh Anh
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Nguyễn Lê Phước Thịnh
16 tháng 5 2022 lúc 7:06

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)

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

Nguyễn Đức Hoàn
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Nguyễn Lê Phước Thịnh
12 tháng 1 2023 lúc 9:24

a: ĐKXĐ: x<>1; x<>2; x<>3

\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)

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

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

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

b:

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Hoàng Huy
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Nguyễn Lê Phước Thịnh
28 tháng 7 2021 lúc 22:05

a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)

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

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

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

\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)

b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)

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

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

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