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![](https://rs.olm.vn/images/avt/0.png?1311)
a,A = \(\dfrac{3}{x-1}\)
A \(\in\) Z \(\Leftrightarrow\) 3 ⋮ \(x-1\) ⇒ \(x-1\) \(\in\) { -3; -1; 1; 3}
\(x\) \(\in\) { -2; 0; 2; 4}
b, B = \(\dfrac{x-2}{x+3}\)
B \(\in\) Z \(\Leftrightarrow\) \(x-2\) \(⋮\) \(x+3\) ⇒ \(x+3-5\) \(⋮\) \(x+3\)
⇒ 5 \(⋮\) \(x+3\)
\(x+3\) \(\in\){ -5; -1; 1; 5}
\(x\) \(\in\) { -8; -4; -2; 2}
a.\(A=\dfrac{3}{x-1}\)có giá trị là 1 số nguyên khi \(3\) ⋮ \(x-1.\)
\(\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}.\)
Ta có bảng:
\(x-1\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
\(x\) | \(2\) | \(0\) | \(4\) | \(-2\) |
TM | TM | TM | TM |
Vậy \(x\in\left\{-2;0;2;4\right\}.\)
b.\(B=\dfrac{x-2}{x+3}\)có giá trị là 1 số nguyên khi \(x-2\) ⋮ \(x+3.\)
\(\Rightarrow\left(x+3\right)-5⋮x+3.\)
Mà x+3 ⋮ x+3 \(\Rightarrow\) Ta cần: \(-5⋮x+3\Rightarrow x+3\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}.\)
Ta có bảng:
\(x+3\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
\(x\) | \(-2\) | \(-4\) | \(2\) | \(-8\) |
TM | TM | TM | TM |
Vậy \(x\in\left\{-8;-4;-2;2\right\}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Với $a,b,c>0$ ta có:
$M> \frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}=\frac{a+b+c}{a+b+c}{a+b+c}=1(*)$
Mặt khác:
Xét hiệu: $\frac{a}{a+b}-\frac{a+c}{a+b+c}=\frac{-bc}{(a+b)(a+b+c)}<0$ với mọi $a,b,c>0$
$\Rightarrow \frac{a}{a+b}< \frac{a+c}{a+b+c}$
Tương tự ta cũng có: $\frac{b}{b+c}< \frac{b+a}{a+b+c}; \frac{c}{c+a}< \frac{c+b}{a+b+c}$
Cộng lại ta được: $M< \frac{a+c+b+a+c+b}{a+b+c}=\frac{2(a+b+c)}{a+b+c}=2(**)$
Từ $(*); (**)\Rightarrow 1< M< 2$ nên $M$ không là số nguyên.
![](https://rs.olm.vn/images/avt/0.png?1311)
`A = (n+3)/(n-2)`
Ta có:
`(n+3)/(n-2)`
`=> (n+3)/(n+3-5)`
`=> -5 : n+3` hay `n+3 in Ư(-5)`
Biết: `Ư(-5)={-1;1;-5;5}`
`=> n in{-3;1;3;7}`
Ta có:
n + 3 = n - 2 + 5
Để A ∈ Z thì n - 2 ∈ Ư(5) = {-5; -1; 1; 5}
⇒ n ∈ {-3; 1; 3; 7}
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)
\(\Rightarrow A=\frac{20}{21}\)
\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)
Do đó $A>B$
![](https://rs.olm.vn/images/avt/0.png?1311)
A.2=4/1.5+6/5.11+...+12/29.41
A.2=1-1/5+1/5-1/11+...+1/29-1/41
A.2=1-1/41
A.2=40/41
A=20/41
B.3=3/1.4+6/4.10+...+12/29.31
B.3=1-1/4+1/4-1/10+...+1/29-1/31
B.3=1-1/31
B.3=30/31
B=10/31
Vì 20/41.10/31 nên A>B
\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(\Rightarrow2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)
\(\Rightarrow2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)
\(\Rightarrow2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)
\(\Rightarrow A=\dfrac{40}{41}:2=\dfrac{20}{41}\)(1)
\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(\Rightarrow3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{31}=\dfrac{30}{31}\)
\(\Rightarrow B=\dfrac{30}{31}:3=\dfrac{10}{31}\)
\(\Rightarrow B=\dfrac{2}{2}.\dfrac{10}{31}=\dfrac{20}{62}\)
+)Ta có:\(\dfrac{20}{62}< \dfrac{20}{41}\Rightarrow B< A\)
Hay A>B(ĐPCM)
Chúc bn học tốt
Lời giải:
Ta có: \(A=\frac{a}{3}+\frac{a^2}{2}+\frac{a^3}{6}\)
\(\Leftrightarrow A=\frac{2a+3a^2+a^3}{6}\)
Xét tử số:
\(a^3+3a^2+2a=a(a^2+3a+2)\)
\(=a[a(a+2)+(a+2)]\)
\(=a(a+1)(a+2)\)
Vì $a,a+1$ là hai số nguyên liên tiếp nên
\(a(a+1)\vdots 2\Rightarrow a(a+1)(a+2)\vdots 2\)
\(\Leftrightarrow a^3+3a^2+2a\vdots 2\) (1)
Mặt khác \(a,a+1,a+2\) là ba số nguyên liên tiếp nên tích của chúng chia hết cho $3$
\(\Leftrightarrow a(a+1)(a+2)\vdots 3\)
\(\Leftrightarrow a^3+3a^2+2a\vdots 3\) (2)
Từ (1)(2) kết hợp với $(2,3)$ nguyên tố cùng nhau suy ra \(a^3+3a^2+2a\vdots 6\)
\(\Rightarrow A=\frac{a^3+3a^2+2a}{6}\in\mathbb{Z}\). Ta có đpcm.