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\(\dfrac{1}{a}-\dfrac{1}{b}=1\)
\(\Leftrightarrow\dfrac{b-a}{ab}=1\)
\(\Leftrightarrow b-a=ab\)
\(\Leftrightarrow a+ab-b=0\)
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)
\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)
\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)
\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)
\(\Leftrightarrow-x=7\)
\(\Leftrightarrow x=-7\)
\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)
Theo đề, ta có: \(\dfrac{1+2x}{18}=\dfrac{1+4x}{34}\)
\(\Leftrightarrow34\left(1+2x\right)=18\left(1+4x\right)\)
\(\Leftrightarrow34+68x=18+72x\)
\(\Leftrightarrow34-18=72x-68x\)
\(\Leftrightarrow16=4x\)
\(\Leftrightarrow x=4\)
Khi \(x=4\) vào ta có: \(\dfrac{1+4.4}{34}=\dfrac{1+6.4}{2y^2}\Leftrightarrow\dfrac{1}{2}=\dfrac{25}{2y^2}\)
\(\Leftrightarrow2y^2=50\)
\(\Leftrightarrow y^2=50\)
\(\Leftrightarrow y=\pm5\)
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561
