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e: \(=\dfrac{-3}{4}+\dfrac{5}{7}=\dfrac{-21+20}{28}=\dfrac{-1}{28}\)
a ) \(\dfrac{6}{13}\) + \(\dfrac{-14}{39}\)
= \(\dfrac{6.3}{13.3}\) + \(\dfrac{-14}{39}\)
= \(\dfrac{18}{39}\) - \(\dfrac{14}{39}\)
= \(\dfrac{4}{39}\)
{ các ý còn lại tương tự }
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Thầy giáo dạy phải biết yêu nước thương nòi.
Ai cũng phải nhớ về quê cha đắt tổ.
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\(a,x\ge2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(x-2\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-2x+4\\ P=-5x+\dfrac{37}{8}\)
\(b,x< 2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(2-x\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-4+2x\\ P=-x-\dfrac{27}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{11}{125}-\left(\dfrac{11}{18}-\dfrac{4}{9}\right)-\left(\dfrac{5}{7}-\dfrac{17}{14}\right)\\ =\dfrac{11}{125}-\dfrac{1}{3}+\dfrac{1}{2}\\ =\dfrac{11}{125}+\dfrac{1}{6}=\dfrac{191}{750}\)
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Bài 2:
1: ĐKXĐ: x<>1
\(\dfrac{x}{x-1}+\dfrac{1}{1-x}\)
\(=\dfrac{x}{x-1}-\dfrac{1}{x-1}\)
\(=\dfrac{x-1}{x-1}=1\)
2: ĐKXĐ: x<>3/2
\(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)
\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)
\(=\dfrac{11x+x-18}{2x-3}=\dfrac{12x-18}{2x-3}\)
\(=\dfrac{6\left(2x-3\right)}{2x-3}\)
=6
3: ĐKXĐ: x<>1/2
\(\dfrac{4x+5}{2x-1}+\dfrac{5-9x}{1-2x}\)
\(=\dfrac{4x+5}{2x-1}+\dfrac{9x-5}{2x-1}\)
\(=\dfrac{4x+5+9x-5}{2x-1}=\dfrac{13x}{2x-1}\)
4: ĐKXĐ: x<>2/5
\(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)
\(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)
5: ĐKXĐ: \(x\ne\pm y\)
\(\dfrac{xy}{x^2-y^2}-\dfrac{x^2}{y^2-x^2}\)
\(=\dfrac{xy}{x^2-y^2}+\dfrac{x^2}{x^2-y^2}\)
\(=\dfrac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x-y}\)
6: ĐKXĐ: \(x\notin\left\{0;7\right\}\)
\(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)
\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)
\(=\dfrac{5x-35}{x\left(5x-35\right)}=\dfrac{1}{x}\)
7: ĐKXĐ: \(x\ne1\)
\(\dfrac{x+2}{x-1}-\dfrac{x-9}{1-x}-\dfrac{x-9}{1-x}\)
\(=\dfrac{x+2}{x-1}+\dfrac{x-9}{x-1}+\dfrac{x-9}{x-1}\)
\(=\dfrac{x+2+x-9+x-9}{x-1}=\dfrac{3x-16}{x-1}\)
8: ĐKXĐ:x<>1
\(\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
\(=\dfrac{2x^2-x}{x-1}-\dfrac{x+1}{x-1}+\dfrac{2-x^2}{x-1}\)
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}\)
=x-1
9: ĐKXĐ: x<>3
\(\dfrac{4-x^2}{x-3}+\dfrac{2x-x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}+\dfrac{x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2+x^2-2x+5-4x}{x-3}=\dfrac{-6x+9}{x-3}\)
10: ĐKXĐ: x<>5
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{5-x}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1}{x-5}-\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1-x+18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>1/5 so cam la 6 qua
=> so cam la 30 qua
=>so quyt la 5 qua
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88+220=(23)8+220=224+220=224(216+1)=224x17chia het cho 17
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