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Nguyễn Ngọc Thanh
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Lê Thanh Ngọc
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Nguyễn Ngọc Phượng
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Võ Đông Anh Tuấn
8 tháng 7 2016 lúc 11:25

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)

\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

Quốc An
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Võ Đông Anh Tuấn
7 tháng 8 2016 lúc 8:50

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)

\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

Nguyễn Hải Anh Jmg
7 tháng 8 2016 lúc 9:03

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
\(=\frac{1}{\left(x+3\right)^2}-\frac{1}{\left(x-3\right)^2}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}-\frac{\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\frac{x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)
 

Kuramajiva
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Nguyễn Việt Lâm
12 tháng 7 2021 lúc 22:02

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

Nguyễn Việt Lâm
12 tháng 7 2021 lúc 22:04

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

Nguyễn Việt Lâm
12 tháng 7 2021 lúc 22:07

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Lê Ly
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Quỳnh Anh
21 tháng 7 2021 lúc 10:29

Trả lời:

a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)

d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)

e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)

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Đoàn Tuấn Anh
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❤  Hoa ❤
1 tháng 12 2018 lúc 19:44

\(C=\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}.\)

\(C=\frac{1}{\left(x+3\right)^2}+\frac{-1}{-\left(6x-x^2-9\right)}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)

\(C=\frac{1}{\left(x+3\right)^2}+\frac{-1}{-6x+x^2+9}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)

\(C=\frac{x-3}{\left(x+3\right)\left(x-3\right)}+\frac{-\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)

\(C=\frac{x-3.-x-3.x}{\left(x+3\right).\left(x-3\right)}=\frac{-6x}{\left(x+3\right)\left(x-3\right)}=\frac{-6x}{\left(x^2-9\right)}\)

Ngọc Huyền
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Huỳnh Nguyễn Tuấn Nam
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