x+2023/6+x+2023/12+x+2023/20 +...+x+2023/9900=49
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12 tháng 8 2023
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
12 tháng 8 2023
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
NT
=
3
VQ
26 tháng 5 2022
2023×6+7×2023−2023
=2023×6+7×2023−2023x1
=2023×(6+7−1)
=2023×12
=24276
HC
1
31 tháng 7 2023
- Với \(0< x;y< 1\)
\(x^2>x^{2003}\left(1\right)\)
\(y^2>y^{2003}\left(2\right)\)
\(z^2>z^{2003}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow M=x^2+y^2+z^2>x^{2003}+y^{2003}+z^{2003}=3\)
\(\Rightarrow\) Không có giá trị max của M.
- Với \(x;y\ge1\)
\(x^2\le x^{2003}\left(1\right)\)
\(y^2\le y^{2003}\left(2\right)\)
\(z^2\le z^{2003}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow x^2+y^2+z^2\le x^{2003}+y^{2003}+z^{2003}=3\)
\(\Rightarrow Max\left(M\right)=3\left(x=y=z=1\right)\)
T
22 tháng 12 2023
Ta có:
\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)
nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)
Thay \(x=4;y=1\) vào \(P\), ta được:
\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)
\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)
\(=1-1=0\)
Vậy \(P=0\) khi \(x=4;y=1\).
13 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`2023 \times 14 + 2023 \times 83+97`
`= 2023 \times (14+83) + 97`
`= 2023 \times 97 + 97`
`= 196231 +97`
`= 196328`
`78 \times 31+78 \times 24+78 \times 17+78 \times 22 \times 72`
`= 78 \times (31+24+17+22 \times 72)`
`= 78 \times (31+24+17+1584)`
`= 78 \times 1656`
`= 129168`
`371 \times 69+371 \times 14+629 \times 83`
`= 371 \times (69+14) + 629 \times 83`
`= 371 \times 83 + 629 \times 83`
`= 83 \times (371+629)`
`= 83 \times 1000`
`= 83000`
`91 \times 51+49 \times 163-49 \times 72`
`= 91 \times 51 + 49 \times (163-72)`
`= 91 \times 51+49 \times 91`
`= 91 \times (51+49)`
`= 91 \times 100`
`= 9100`
`3 \times 12 \times 17 + 4 \times 9 \times 81 + 2 \times 6 \times 6`
`= 9 \times 4 \times 17 + 4 \times 9 \times 81 + 4 \times 9 \times 2`
`= 9 \times 4 \times (17+81+2)`
`= 36 \times 100`
`= 3600`
`42 \times 17 + 58 \times 83 \times 42 \times 83 + 42 \times 83 + 17`
`= 17 \times (42+58) + 42 \times 83 \times (58 \times 83 +1)`
`= 17 \times 100 + 3486 \times 4815`
`= 1700 + 16785090`
`= 16786790`
HD
5
17 tháng 9 2023
\(2023\times28+2023\times34-2023\times52\)
\(=2023\times\left(28+34-52\right)\)
\(=2023\times10\)
\(=20230\)
17 tháng 9 2023
`# \text {DNamNgV}`
`2023 \times 28 + 2023 \times 34 - 2023 \times 52`
`= 2023 \times (28 + 34 - 52)`
`= 2023 \times 10 `
`=20230`
DT
2
28 tháng 2 2023
\(y\times2023-y=2023\times2021+2023\)
\(y\times\left(2023-1\right)=2023\times\left(2021+1\right)\)
\(y\times2022=2023\times2022\)
\(y=2023\times2022\div2022\)
\(y=2023\)
D
3
\(\dfrac{x+2023}{6}+\dfrac{x+2023}{12}+\dfrac{x+2023}{20}+...+\dfrac{x+2023}{9900}=49\)
\(\left(x+2023\right)\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)=49\)
\(\left(x+2023\right)\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\right)=49\)
\(\left(x+2023\right)\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=49\)
\(\left(x+2023\right)\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=49\)
\(\left(x+2023\right).\dfrac{49}{100}=49\)
\(x+2023=49:\dfrac{49}{100}\)
\(x+2023=100\)
\(x=100-2023\)
\(x=-1923\)
x+2023/6+x+2023/12+x+2023/20 +...+x+2023/9900=49
(x+2023). (1/6+1/12+1/20+...+1/9900)=49
(x+2023).(1/2.3+1/3.4+1/4.5+...+1/99.100)=49
(x+2023).(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)=49
(x+2023).(1/2-1/100) =49
*còn lại biết tính rồi đoá* mình hỏi với làm cho vui để ôn thôi , các bạn nào ko bt làm thì có thể tham khảo ak