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TL
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H
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25 tháng 2 2020
1) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}\)
\(=\frac{-4x^2+8x-4}{-4x^3+4x^2+4x-4}\)
\(=\frac{-x^2+2x-1}{-x^3+x^2+x-1}\)
\(=\frac{\left(-x+1\right)\left(x-1\right)}{\left(-x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(=\frac{1}{x+1}\)
2) \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
\(=\frac{-16x^3+16x^2-4x}{-16x^4+16x^3-4x^2}\)
\(=\frac{-16x^2+16x-4}{-16x^3+16x^2-4x}\)
\(=\frac{-4x^2+4x-1}{-4x^3+4x^2-x}\)
\(=\frac{\left(-2x+1\right)\left(2x-1\right)}{x\left(-2x+1\right)\left(2x-1\right)}\)
\(=\frac{1}{x}\)
20 tháng 11 2018
\(\frac{2x+y}{2x^2-xy}+\frac{8y}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
\(=\frac{2x+y}{x\left(2x-y\right)}-\frac{8y}{\left(2x-y\right)\left(2x+y\right)}+\frac{2x-y}{x\left(2x+y\right)}\)
\(=\frac{\left(2x+y\right)^2-8xy+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{8x^2-8xy+2y^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(4x^2-4xy+y^2\right)}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{2\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
BH
16 tháng 8 2019
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
VM
1 tháng 2 2020
Bài 4:
a) \(\frac{2x^2-10xy}{2xy}+\frac{5y-x}{y}\)
\(=\frac{y.\left(2x^2-10xy\right)}{2xy.y}+\frac{2xy.\left(5y-x\right)}{2xy.y}\)
\(=\frac{2x^2y-10xy^2}{2xy^2}+\frac{10xy^2-2x^2y}{2xy^2}\)
\(=\frac{2x^2y-10xy^2+10xy^2-2x^2y}{2xy^2}\)
\(=\frac{0}{2xy^2}\)
\(=0.\)
b) \(\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{x^2-y^2}\)
\(=\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{1.\left(x+y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y}{\left(x-y\right).\left(x+y\right)}+\frac{x+y}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y+x+y+3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x-y}{\left(x-y\right).\left(x+y\right)}\)
c) \(x+y+\frac{x^2+y^2}{x+y}\)
\(=\frac{x+y}{1}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right).\left(x+y\right)}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right)^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2+x^2+y^2}{x+y}\)
\(=\frac{2x^2+2xy+2y^2}{x+y}.\)
Chúc bạn học tốt!
HN
16 tháng 12 2018
\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)
\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)
\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)
\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)
\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)
\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)
kết quả này có đúng không thì mình chưa chắc bạn nhé : \(\frac{4x+16}{y^2-4x}\)