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mạnh
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Hoàng Thị Lan Hương
26 tháng 7 2017 lúc 14:56

Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)

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

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

\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)

ミ★Ƥɦươŋɠ Ňɦї★彡
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ĐKx\(\ne\)2,x\(\ne\)0

\(=\)\(\frac{2(x+2)+2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\):\(\frac{4x}{\left(x+2\right)^2}\)

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

=\(\frac{x+2}{x-2}\)

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KAl(SO4)2·12H2O
28 tháng 12 2019 lúc 15:07

\(\left(\frac{2}{x-2}+\frac{2}{x+2}\right):\frac{4x}{x^2+4x+4}\)

\(=\left(\frac{2}{x-2}+\frac{2}{x+2}\right):\frac{4x}{\left(x+2\right)^2}\)

\(=\left(\frac{2}{x-2}+\frac{2}{x+2}\right).\frac{\left(x+2\right)^2}{4x}\)

\(=\frac{4x}{x^2-4}.\frac{\left(x+2\right)^2}{4x}\)

\(=\frac{4x.\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right).4}\)

\(=\frac{x+2}{x-2}\)

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Đỗ Phương Thảo
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Chu Công Đức
20 tháng 2 2020 lúc 9:16

\(ĐKXĐ:x\ne\pm2\)

\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(=\left[\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right]\)

\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)

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

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

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Bùi Anh Tuấn
20 tháng 2 2020 lúc 9:17

\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(\Leftrightarrow\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)

\(\Leftrightarrow\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2x}{\left(x+2\right)^2}\cdot\frac{\left(x-2\right)\left(x+2\right)}{x+4}\)

\(\Leftrightarrow\frac{2x^2-4x}{\left(x+2\right)\left(x+4\right)}\)

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Dư Hạ Băng
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Kim Trân Ni
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KAl(SO4)2·12H2O
9 tháng 12 2019 lúc 14:47

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

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

\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)

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

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

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KAl(SO4)2·12H2O
9 tháng 12 2019 lúc 14:51

\(\frac{x^2+4x+4}{1-x}.\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\)

\(=\frac{x^2+4x+4}{1-x}.\left[\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\right]\)

\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)

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

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

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Yukihira Souma
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Phạm Hồ Thanh Quang
15 tháng 7 2017 lúc 7:46

a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)

b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)

gh
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๖²⁴ʱ๖ۣۜTɦủү❄吻༉
22 tháng 4 2020 lúc 15:33

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

\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)

\(x\left(x+1\right)+x\left(x-3\right)=4x\)

\(x^2+x+x^2-3x=4x\)

\(2x^2-2x=4x\)

\(2x^2-2x-4x=0\)

\(2x\left(x-3\right)=0\)

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

hoặc 

\(x-3=0\Leftrightarrow x=3\)

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Nobi Nobita
22 tháng 4 2020 lúc 16:00

b) \(ĐKXĐ:x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )

Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)

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nguyenthithuhien
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Nguyễn Thanh Thảo
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