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a) \(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a-b}\left(đk:a,b\ne0,a\ne b\right)\Leftrightarrow\dfrac{b-a}{ab}=\dfrac{1}{a-b}\)
\(\Leftrightarrow-\left(a-b\right)^2=ab\Leftrightarrow a^2-ab+b^2=0\)
\(\Leftrightarrow\left(a^2-ab+\dfrac{1}{4}b^2\right)+\dfrac{3}{4}b^2=0\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-\dfrac{1}{2}b=0\\\dfrac{3}{4}b^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}b\\b=0\end{matrix}\right.\) \(\Leftrightarrow a=b=0\left(ktm\right)\)
Vậy k có a,b thõa mãn
b) \(\dfrac{5}{2a}=\dfrac{1}{6}+\dfrac{b}{3}\left(a\ne0\right)\Leftrightarrow\dfrac{2b+1}{6}-\dfrac{5}{2a}=0\Leftrightarrow\dfrac{a\left(2b+1\right)-15}{6a}=0\)
\(\Leftrightarrow a\left(2b+1\right)-15=0\Leftrightarrow a\left(2b+1\right)=15\)
Do \(a,b\in Z,a\ne0\) nên ta có bảng sau:
a | 1 | -1 | 15 | -15 | 3 | -3 | 5 | -5 |
2b+1 | 15 | -15 | 1 | -1 | 5 | -5 | 3 | -3 |
b | 7(tm) | -8(tm) | 0(tm | -1(tm) | 2(tm) | -3(tm) | 1(tm) | -2(tm) |
Vậy...
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\(a)\) Ta có :
\(A=\frac{6n-2}{3n+1}=\frac{6n+2-4}{3n+1}=\frac{2\left(3n+1\right)-4}{3n+1}=\frac{2\left(3n+1\right)}{3n+1}-\frac{4}{3n+1}=2+\frac{4}{3n+1}\)
Để A là số nguyên thì \(\frac{4}{3n+1}\) phải là số nguyên \(\Rightarrow\)\(4⋮\left(3n+1\right)\)\(\Rightarrow\)\(\left(3n+1\right)\inƯ\left(4\right)\)
Mà \(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Do đó :
\(3n+1\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(4\) | \(-4\) |
\(n\) | \(0\) | \(\frac{-2}{3}\) | \(\frac{1}{3}\) | \(-1\) | \(1\) | \(\frac{-5}{3}\) |
Lại có \(n\inℤ\) nên \(n\in\left\{-1;0;1\right\}\)
Câu b) là tương tự rồi tính n ra, sau đó thấy n nào giống với câu a) rồi trả lời
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+) Ta có: 2 . (a - 1) \(\le\) 0 \(\Rightarrow\) a - 1 \(\le\) 0 \(\Rightarrow\) |a - 1| = -(a - 1)
+) Ta có:
P = (a - 1) - (a - 1) + (a - 1) - (a - 1) + ... + (a - 1) - (a - 1)
P = 0 + 0 + ... + 0
P = 0
\(2\left(a-1\right)\le0\Leftrightarrow\left(a-1\right)\le0\Leftrightarrow a\le1\Leftrightarrow\left|a-1\right|=-\left(a-1\right)=1-a\)
\(\Rightarrow P=\left(a-1\right)+\left|a-1\right|+\left(a-1\right)+\left|a-1\right|+.........++\left(a-1\right)+\left|a-1\right|\left(\text{100 số hạng}\right)=\left(a-1+1-a\right)+\left(a-1+1-a\right)+........+\left(a-1+1-a\right)\left(\text{50 số hạng}\right)=0+0+.....+0=0.\text{ Nên: P=0}\)
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