NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
BV
0
BV
Tìm số dư trong phép chia (2023\(\left(2023^{2024}+2024^{2025}+2025^{2026}\right)^{10}\)chia cho 111
0
18 tháng 6 2019
\(\frac{2015}{2016}< \frac{2016}{2016}=1=\frac{2034}{2034}< \frac{2035}{2034}\)
\(\Rightarrow\frac{2015}{2016}< \frac{2035}{2034}\)
\(\frac{-2025}{2024}< \frac{-2024}{2024}=-1< \frac{-2026}{2027}\)
\(\Rightarrow\frac{-2025}{2024}< \frac{-2026}{2027}\)
T
18 tháng 6 2019
#)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 \(\frac{-1}{2024}< \frac{-1}{2027}\Rightarrow\frac{-2025}{2024}< \frac{-2026}{2027}\)
HD
3
18 tháng 9 2017
\(\left|x+1\right|+\left|x+2\right|+.........+\left|x+101\right|=2024x\)
\(\Leftrightarrow\left|101x+\dfrac{\left[\left(101-1\right):1+1\right]\left(101+1\right)}{2}\right|=2024x\)
\(\Leftrightarrow\left|101x+5151\right|=2024x\)
\(\Leftrightarrow\left|101x+5151\right|-2024x=0\)
\(\Leftrightarrow-1923x+5151=0\)
\(\Leftrightarrow-1923x=5151\)
\(\Leftrightarrow x=\dfrac{5151}{-1923}\)
Vậy ..
18 tháng 9 2017
đề mình ko ghi lại nhé
\(\Rightarrow\left|101x+\dfrac{\left[\left(101-1\right):1+1\right]\left(101+1\right)}{2}\right|=2024x\)
\(\Rightarrow\left|101x+5151\right|=2024x\)
\(\Rightarrow-1923+5151=0\)
\(\Rightarrow-1923x=5151\Rightarrow x=\dfrac{5151}{-1923}\)
25 tháng 2 2017
\(S=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2025}-\sqrt{2024}}\)
Ta nhận xét thấy mỗi số hạng trong S đều dương. Từ đó ta đặt
\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2024}-\sqrt{2023}}\left(A>0\right)\)
\(\Rightarrow S=A+\frac{1}{\sqrt{2025}-\sqrt{2024}}=A+\frac{\sqrt{2025}+\sqrt{2024}}{\left(\sqrt{2025}-\sqrt{2024}\right)\left(\sqrt{2025}+\sqrt{2024}\right)}\)
\(=A+\sqrt{2025}+\sqrt{2024}>\sqrt{2025}=45\)
Vậy \(S>45\)
PS: Phan Thanh Tịnh xem lại bài giải nhé bạn
24 tháng 2 2017
Ta có : 1 = (n + 1) - n =\(\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\)
\(=\left(\sqrt{n+1}\right)^2-\sqrt{n+1}.\sqrt{n}+\sqrt{n+1}.\sqrt{n}+\left(\sqrt{n}\right)^2\)
\(=\sqrt{n+1}.\left(\sqrt{n+1}-\sqrt{n}\right)+\sqrt{n}.\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(=\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n-1}+\sqrt{n}\right)\)\
\(\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Áp dụng vào bài toán,ta có :
\(S=\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2025}-\sqrt{2024}=\sqrt{2025}\)= 45
Vậy S = 45
12 tháng 1 2018
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
C
24 tháng 8 2018
\(a,\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5\)
\(\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{4}{3}x=5\)
\(\dfrac{4}{3}x=\dfrac{2}{3}-5\)
\(\dfrac{4}{3}x=-\dfrac{13}{3}\)
\(x=-\dfrac{13}{3}:\dfrac{4}{3}\)
\(x=-\dfrac{13}{4}\)
Vậy...............
\(b,\left(x+\dfrac{1}{2}\right)\left(\dfrac{3}{4}-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy................
\(c,\dfrac{2x-1}{-3+2}=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
AT
26 tháng 6 2017
a/ \(\left(x+1\right)\left(x-2\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)
TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)
Vậy.........
b/ \(\left(x-3\right)\left(x-4\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)
TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)
Vậy...............
c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)
\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)
Vậy...............
PL
26 tháng 6 2017
Để ( x + 1 ) ( x - 2 ) < 0
=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2
=> x + 1 dương x + 2 âm
Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2
7 tháng 8 2017
bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
`2024x(x-2024/2025)=0`
`=> 2024x= 0` hoặc `x - 2024/2025 = 0`
`=> x = 0 : 2024` hoặc `x = 0 + 2024/2025`
`=> x = 0` hoặc `x = 2024/2025`
Vậy: `x= 0; x = 2024/2025`