\(\dfrac{2n+5}{n-4}=\dfrac{2n-8+13}{n-4}=\dfrac{2\left(n-4\right)+13}{n-4}=2+\dfrac{13}{n-4}\)

Để \(\dfrac{2n-5}{n-4}\) là số nguyên thì 13 ⋮ n - 4

⇒ n - 4 ∈ Ư(13) = {1; -1; 13; -13}

⇒ n ∈ { 5; 3; 17; -9} 

Các giá trị nguyên của $n$ thỏa mãn điều kiện là $n=-9$ và $n=17$.

Lời giải:

Đặt $M=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$

Với $a,b,c$ nguyên dương thì:

$M=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}> \frac{b}{a+b+c}+\frac{c}{b+c+a}+\frac{a}{c+a+b}=\frac{a+b+c}{a+b+c}=1(*)$
Lại có:

Xét hiệu $\frac{b}{a+b}-\frac{b+c}{a+b+c}=\frac{b(a+b+c)-(a+b)(b+c)}{(a+b)(a+b+c)}$

$=\frac{-b^2}{(a+b)(a+b+c)}<0$ với mọi $a,b,c$ nguyên dương.

$\Rightarrow \frac{b}{a+b}< \frac{b+c}{a+b+c}$
Tương tự: 

$\frac{c}{b+c}< \frac{c+a}{b+c+a}$

$\frac{a}{c+a}< \frac{a+b}{c+a+b}$
$\Rightarrow M< \frac{b+c}{a+b+c}+\frac{c+a}{b+c+a}+\frac{a+b}{c+a+b}=\frac{2(a+b+c)}{a+b+c}=2(**)$
Từ $(*); (**)\Rightarrow 1< M< 2$

Do đó $M$ không phải số nguyên.

Bạn lưu ý lần sau gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

a\(\): \(K=1-5+5^2-5^3+...+5^{100}\)

=>\(5K=5-5^2+5^3-5^4+...+5^{101}\)

=>\(5K+K=5-5^2+5^3-5^4+...+5^{101}+1-5+5^2-5^3+...+5^{100}\)

=>\(6K=5^{101}+1\)

=>\(K=\dfrac{5^{101}+1}{6}\)

b: \(5^{101}\) chia 6 sẽ dư 5 bởi vì \(5^{101}+1⋮6\) và 1+5=6

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012-1}{4048}=\dfrac{1011}{4048}\)

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+\dfrac{1}{4\cdot10}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2023\cdot2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1012-1}{2024}\)

\(=\dfrac{1011}{4048}\)

Có đáp án luôn rồi!

@456 troll thật

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}.2021\)

\(=\left\{47-\left[736:2^4\right]\right\}.2021\)

\(=\left\{47-\left[736:16\right]\right\}.2021\)

\(=\left\{47-46\right\}.2021\)

\(=1.2021\)

\(=2021\)

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}\cdot2021\)

\(=\left\{47-736:16\right\}\cdot2021\)

\(=\left(47-46\right)\cdot2021=2021\)

2(x-1)+3(x-2)=x-4

=> 2x-2+3x-6=x-4

=> 5x-8=x-4

=> 5x-x=8-4

=> 4x=4

=> x=4:4

=> x=1

Vậy: x=1

\(2\left(x-1\right)+3\left(x-2\right)=x-4\)

\(2x-2+3x-6=x-4\)

\(\left(2x+3x\right)-\left(2+6\right)=x-4\)

\(5x-8=x-4\)

\(5x-x=-4+8\)

\(4x=4\)

\(x=1\)

\(\left(2345-45\right)+2345\)

\(=2300+2345\)

\(=4645\)

\(\left(2345-45\right)+2345\)

\(=2300+2345\)

\(=4645\)

\(\left(7+x\right)-\left(21-13\right)=32\)

\(\left(7+x\right)-8=32\)

\(7+x=32+8\)

\(7+x=40\)

\(x=40-7\)

\(x=33\)

Vậy \(x=33\)

\(\left(7+x\right)-\left(21-13\right)=32\)

\(7+x-21+13=32\)

\(x+\left(7+13-21\right)=32\)

\(x-1=32\)

\(x=33\)

\(-11-\left(19-x\right)=50\)

\(-11-19+x=50\)

\(x-\left(11+19\right)=50\)

\(x-30=50\)

\(x=50+30\)

\(x=80\)

\(-11-\left(19-x\right)=50\Leftrightarrow19-x=-11-50=-61\Leftrightarrow x=19+61=80\)