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Đặt 3154=a; 6517=b
Theo đề, ta có:
\(B=2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{3}{a}\cdot\dfrac{b-1}{b}+\dfrac{6}{\dfrac{1}{2}a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{2a+1}{a}\cdot\dfrac{3}{b}-\dfrac{3b-3}{ab}+\dfrac{12}{a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{6a+3-3b+3-6}{ab}+\dfrac{12b}{ab}\)
\(=\dfrac{6a-3b+12b}{ab}=\dfrac{6a+9b}{ab}=\dfrac{3\left(2a+3b\right)}{ab}\)
\(=\dfrac{3\left(2\cdot3154+3\cdot6517\right)}{3154\cdot6517}\)
a) \(\frac{5}{6}=\frac{-1}{x}\) \(\Leftrightarrow\) 5x = -1 . 6 \(\Leftrightarrow\) x = \(\frac{-6}{5}\) = -1,2
b)\(\frac{-3}{7}=\frac{x}{14}\Leftrightarrow\) -3 . 14 = 7x \(\Leftrightarrow\) x = \(\frac{-3.14}{7}\) = -6
c)\(\frac{x+1}{4}=\frac{-3}{2}\) \(\Leftrightarrow\) 2(x+1) = -3.4\(\Leftrightarrow\) \(x+1=\frac{-3.4}{2}\) = -6 \(\Leftrightarrow\) x = -6 - 1= -7
d) \(\frac{2x-3}{5}=\frac{-6}{7}\) \(\Leftrightarrow\) 2x - 3 = \(\frac{-6.5}{7}\) = \(\frac{-30}{7}\) \(\Leftrightarrow\) 2x = \(\frac{-30}{7}+3\) = \(\frac{-9}{7}\)
\(\Leftrightarrow\) x = \(\frac{-9}{7}:2\) = \(\frac{-9}{14}\)
e) \(\frac{3-5x}{4}=\frac{5}{6}\) \(\Leftrightarrow\) 3-5x = \(\frac{5.4}{6}\) = \(\frac{10}{3}\) \(\Leftrightarrow\) 5x = 3-\(\frac{10}{3}\) = \(\frac{-1}{3}\) \(\Leftrightarrow\) x = \(\frac{-1}{15}\)
f)\(\frac{12}{x}=\frac{-6}{5}\) \(\Leftrightarrow\) x = \(\frac{12.5}{-6}\) = -10
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
c) \(\frac{x+1}{3}>\frac{2x-1}{6}-2\)
⇔\(\frac{2\left(x+1\right)}{6}>\frac{2x-1-12}{6}\)
⇔2x + 2 > 2x - 13
⇔0x > -15
Vậy S=Φ
b) \(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)
⇔\(\frac{x-2-2\left(x-1\right)}{6}\le\frac{3x}{6}\)
⇔x - 2 - 2x +2 ≤ 3x
⇔-4x ≤ 0
⇔x ≥ 0
Vậy S={x | x ≥ 0}
1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)= \(\frac{6x}{3xy}\)=\(\frac{3}{y}\)
2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1
3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)
4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)
5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)
=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)