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LT
2
AH
Akai Haruma
Giáo viên
13 tháng 1 2024
Lời giải:
Ta thấy: $(x-2022)^2\geq 0$ với mọi $x$
$\Rightarrow (x-2022)^2+2\geq 2$
$\Rightarrow \frac{6}{(x-2022)^2+2}\leq 3$ với mọi $x$ (1)
$|y-2023|\geq 0$ với mọi $y$
$\Rightarrow |y-2023|+3\geq 3$ với mọi $y$ (2)
Từ (1); (2) suy ra để $\frac{6}{(x-2022)^2+2}=|y-2023|+3$ thì:
$\frac{6}{(x-2022)^2+2}=|y-2023|+3=3$
$\Rightarrow x-2022=y-2023=0$
$\Leftrightarrow x=2022; y=2023$
BB
2
AH
Akai Haruma
Giáo viên
31 tháng 1 2024
Lời giải:
\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)
\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)
\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)
\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)
HN
2
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$
$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$
$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$
$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$
$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$
DD
4 tháng 3 2024
A=23.41+24.51+....+22023.20241
=23.4+24.5+...+22023.2024=3.42+4.52+...+2023.20242
=2(4−33.4+5−44.5+...+2024−20232023.2024)=2(3.44−3+4.55−4+...+2023.20242024−2023)
=2(13−14+14−15+....+12023−12024)=2(31−41+41−51+....+20231−20241)
=2(13−12024)=20213036=2(31−20241)=30362021
TG
1
12 tháng 7 2023
a: =-3/4-1/4+2/7+5/7+2023/2024
=-1+1+2023/2024=2023/2024
b: 2/3x=2/7
=>x=2/7:2/3=3/7
c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
=>x=3/5:2/3=3/5*3/2=9/10
NB
15 tháng 6 2023
\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+\dfrac{15}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)
\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{2023^2}\)
\(A=(1+1+1+...+1)-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{2023^2})\)
Tổng số hạng của 2 ngoặc trên bằng nhau và =(2023-2):1+1=2022(số hạng)
\(A=2022-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2})\)
Ta thấy:
\(0<\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)
Ta có
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}<1\)
Do đó,2021<A<2022
Vậy giá trị của A không phải 1 số tự nhiên(đpcm)
H9
HT.Phong (9A5)
CTVHS
16 tháng 8 2023
\(S=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow\dfrac{25}{5}=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\)
\(\Rightarrow5S+S=\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\right)\)
\(\Rightarrow6S=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow6S=-1-\dfrac{1}{5^{2023}}\)
\(\Rightarrow S=\dfrac{-1-\dfrac{1}{5^{2023}}}{6}\)
HT
20 tháng 3 2023
A=322+832+1542+....+20232−120232�=322+832+1542+....+20232-120232
A=1−122+1−132+1−142+....+1−120232�=1-122+1-132+1-142+....+1-120232
A=2022−(122+132+142+...+120232)�=2022-(122+132+142+...+120232)
122+132+142+...+120232<11.2+12.3+13.4+...+12022.2023122+132+142+...+120232<11.2+12.3+13.4+...+12022.2023
11.2+12.3+13.4+...+12022.2023=1−12+12−13+....−1202311.2+12.3+13.4+...+12022.2023=1-12+12-13+....-12023
⇒0<122+132+142+...+120232<1−12023<1⇒0<122+132+142+...+120232<1-12023<1
⇒2022−(122+132+142+...+120232)⇒2022-(122+132+142+...+120232)ko phải số tự nhiên
⇒A⇒� ko phải số tự nhiên
VT
9 tháng 4 2023
322+832+1542+....+20232-120232"" id="MathJax-Element-1-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-table; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=322+832+1542+....+20232−120232�=322+832+1542+....+20232-120232A=
1-122+1-132+1-142+....+1-120232"" id="MathJax-Element-2-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=1−122+1−132+1−1(2+....+1)120232�=1-122+1-132+1-142+....+1-1202321+12+13+...+122023−1
2022-(122+132+142+...+120232)"" id="MathJax-Element-3-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=2022−(122+132+142+...+120232)�=2022-(122+132+142+...+120232)A
122+132+142+.... <20232
Đặt \(A=2+2^3+...+2^{2023}\)
=>\(4A=2^3+2^5+2^7+...+2^{2025}\)
=>\(4A-A=2^3+2^5+...+2^{2025}-2-2^3-...-2^{2023}\)
=>\(3A=2^{2025}-2\)
=>\(A=\dfrac{2^{2025}-2}{3}\)