Đề có phải là:

\(\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}=4\text{ ?}\)

\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-4=0\)

\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-1-1-1-1=0\)

\(\Rightarrow\left(\dfrac{x+1}{2024}-1\right)+\left(\dfrac{x+2}{2025}-1\right)+\left(\dfrac{x+3}{2026}-1\right)+\left(\dfrac{x+4}{2027}-1\right)=0\)

\(\Rightarrow\left(\dfrac{x+1-2024}{2024}\right)+\left(\dfrac{x+2-2025}{2025}\right)+\left(\dfrac{x+3-2026}{2026}\right)+\left(\dfrac{x+4-2027}{2027}\right)=0\)

\(\Rightarrow\dfrac{x-2023}{2024}+\dfrac{x-2023}{2025}+\dfrac{x-2023}{2026}+\dfrac{x-2023}{2027}=0\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\right)=0\)

Mà \(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\ne0\)

\(\Rightarrow x-2023=0\)

\(\Rightarrow x=0+2023\)

\(\Rightarrow x=2023\)

Vậy, \(x=2023.\)

\(\frac{2015}{2016}< \frac{2016}{2016}=1=\frac{2034}{2034}< \frac{2035}{2034}\)

\(\Rightarrow\frac{2015}{2016}< \frac{2035}{2034}\)

\(\frac{-2025}{2024}< \frac{-2024}{2024}=-1< \frac{-2026}{2027}\)

\(\Rightarrow\frac{-2025}{2024}< \frac{-2026}{2027}\)

#)Giải :

a) Ta có :

\(1-\frac{2015}{2016}=\frac{1}{2016}\)

\(1-\frac{2035}{2036}=\frac{1}{2036}\)

Vì \(\frac{1}{2016}>\frac{1}{2036}\Rightarrow\frac{2015}{2016}>\frac{2035}{2036}\)

b) Ta có : 

\(1+\frac{-2025}{2024}=\frac{-1}{2024}\)

\(1+-\frac{2026}{2027}=\frac{-1}{2027}\)

Vì \(\frac{-1}{2024}< \frac{-1}{2027}\Rightarrow\frac{-2025}{2024}< \frac{-2026}{2027}\)

\(a,\frac{x}{x-3}-\frac{6}{x^2-9}=\frac{1}{x+3}\) (đkxđ: x khác 3, -3)

\(\frac{x\left(x+3\right)-6}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)

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

\(x^2+2x-3=0\)

\(\left(x+3\right)\left(x-1\right)=0\)

\(\Longrightarrow\left[\begin{array}{l}x=-3\left(L\right)\\ x=1\left(N\right)\end{array}\right.\)

\(b,\frac{x^2}{x-2}+\frac{x}{1-x}=\frac{4}{x^2-3x+2}\) (đkxđ: \(x\ne1,x\ne2)\)

\(\frac{x^2}{x-2}-\frac{x}{x-1}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(\frac{x^2\left(x-1\right)-x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(x^2\left(x-1\right)-x\left(x-2\right)=4\)

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

\(x^3-2x^2+2x-4=0\)

\(\left(x^3-2x^2\right)+\left(2x-4\right)=0\)

\(x^2\left(x-2\right)+2\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^2+2\right)=0\)

vì \(x^2+2>0\forall x\) ⇒ x - 2 = 0

⇒ x = 2 (ko thoả mãn)

vậy phương trình vô nghiệm

Điều kiện: x khác (-3,-2,1,4)

PT <=> 

\(1+\frac{2}{x-1}+1-\frac{4}{x+2}+1-\frac{6}{x+3}+1+\frac{8}{x-4}=4\)

<=> \(\frac{1}{x-1}-\frac{2}{x+2}-\frac{3}{x+3}+\frac{4}{x-4}=0\)

<=> (x+2)(x+3)(x-4)-2(x-1)(x+3)(x-4)-3(x-1)(x+2)(x-4)+4(x-1)(x+2)(x+3)=0

<=> (x³+x²-14x-24)-2(x³ - 2x²-11x+12) - 3(x³ - 3x²- 6x+8) + 4(x³+4x² + x-6) = 0

<=> x³+x²-14x-24-2x³ + 4x²+22x-24 - 3x³ + 9x²+ 18x-24 + 4x³+16x² + 4x-24 = 0

<=> 30x² + 30x -96=0

<=> 5x² + 5x -16 = 0

Giải ra được: \(\orbr{\begin{cases}x_1=\frac{-5-\sqrt{345}}{10}\\x_2=\frac{-5+\sqrt{345}}{10}\end{cases}}\)

\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

<=> \(\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)

<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0

<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)

<=> x = 2004

Vậy S = {2004}

đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

 \(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)

\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)

\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)

=> x=1

Theo đề bài ta có: \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}-\frac{x-4}{5}-\frac{x-5}{6}>0\)

=> \(\frac{x-1}{2}+1+\frac{x-2}{3}+1+\frac{x-3}{4}+1-\left(\frac{x-4}{5}+1\right)-\left(\frac{x-5}{6}+1\right)>1\)

<=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}>1\)

<=>\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>1\)

<=> \(\left(x+1\right)\cdot\frac{43}{60}>1\)

<=>\(x+1>\frac{60}{43}\)

<=> x>\(\frac{17}{43}\)

Vậy x>17/43

ta nhận xét rằng mỗi số hạng trong tổng \(M\) đều là số dương. Do đó, \(M > 0\).

Áp dụng bất đẳng thức này cho từng số hạng của \(M\), ta có: \(M = \sum_{k = 1}^{2025} \frac{k}{\left(\right. k + 1 \left.\right)^{3}} < \sum_{k = 1}^{2025} \frac{1}{\left(\right. k + 1 \left.\right)^{2}}\)

Đặt \(j = k + 1\). Khi \(k = 1\) thì \(j = 2\), và khi \(k = 2025\) thì \(j = 2026\). Do đó, \(\sum_{k = 1}^{2025} \frac{1}{\left(\right. k + 1 \left.\right)^{2}} = \sum_{j = 2}^{2026} \frac{1}{j^{2}}\).

Giá trị của \(\pi \approx 3.14159\), nên \(\pi^{2} \approx 9.8696\). \(\frac{\pi^{2}}{6} \approx \frac{9.8696}{6} \approx 1.6449\). Vậy \(\sum_{j = 2}^{2026} \frac{1}{j^{2}} < 1.6449 - 1 = 0.6449\).

Do đó, \(M < 0.6449\).

\(=\frac{1}{2^{3}}+\frac{2}{3^{3}}+\frac{3}{4^{3}}+...+\frac{2025}{202 6^{3}}\) \(M > \frac{1}{2^{3}} = \frac{1}{8} = 0.125\)

Ta có \(0.125 < M < 0.6449\). Vì \(M\) nằm trong khoảng \(\left(\right. 0.125 , 0.6449 \left.\right)\), nên \(M\) không thể là một số tự nhiên

Do đó, giá trị của \(M\) không phải là số tự nhiên.

đây mik cx ko chắc chắn lắm