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DN
18 tháng 5 2023
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, \(x=2023.\)
GN
0
NH
0
11 tháng 4 2023
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
| 2x+1 | 1 | 3 | 7 | 21 |
| x | 0 | 1 | 3 | 10 |
| TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
NV
0
11
7
IM
16 tháng 11 2016
a)
\(219-7\left(x+1\right)=100\)
\(\Rightarrow7\left(x+1\right)=119\)
\(\Rightarrow x+1=17\)
=> x = 16
Vậy x = 16
b)
\(\left(3x-6\right)3=3^4\)
\(\Rightarrow\left(x-2\right)3^2=3^4\)
=> x - 2 = 9
=> x = 7
Vậy x = 7
\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2025}{2027}\)
=>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4052}{2027}\)
=>\(2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4052}{2027}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2026}{2027}\)
=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)
=>\(1-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)
=>\(\dfrac{1}{x+1}=\dfrac{1}{2027}\)
=>x+1=2027
=>x=2026