Tìm số tự nhiên a, biết số a chia cho 39 còn dư 37, và khi chia số a cho 13 được
thương bằng số dư.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số có \(5\)chữ số cần tìm là: \(\overline{abcde}\).
Ta có:
\(\overline{abcde2}=3.\overline{2abcde}\)
\(\Leftrightarrow10.\overline{abcde}+2=600000+3.\overline{abcde}\)
\(\Leftrightarrow7.\overline{abcde}=599998\)
\(\Leftrightarrow\overline{abcde}=85714\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(|x-2|=2x-1\)
\(\Rightarrow\)2 trường hợp: \(\orbr{\begin{cases}x-2=2x-1\\-x+2=2x-1\end{cases}}\)
Trường hợp 1: Trường hợp 2:
\(x-2=2x-1\) \(-x+2=2x-1\)
\(x-2x=-1+2\) \(-x-2x=-1-2\)
\(-1x=1\) \(-3x=-3\)
\(x=1\div\left(-1\right)\) \(x=-3\div\left(-3\right)\)
\(x=-1\) \(x=1\)
\(\Rightarrow x=-1\)Hoặc \(x=1\)
Hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^x=2048\)\(\Rightarrow\)\(2^x=2^{11}\)\(\Rightarrow\)\(x=11\)
Cho đề bài tìm y nhé
\(2^x=2048\)
\(\Rightarrow2^x=2^{11}\)
\(x=11\)
\(\text{Vậy }x=11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(0,5.x+\frac{2}{3}.x-x=-4\)
\(0,5.x+\frac{2}{3}.x-x.1=-4\)
\(\left(0,5+\frac{2}{3}-1\right).x=-4\)
\(\frac{1}{6}.x=-4\)
\(x=-4\div\frac{1}{6}\)
\(x=-24\)
\(0,5.x+\frac{2}{3}.x-x=-4\)
\(\Rightarrow0,5.x+\frac{2}{3}.x-x.1=-4\)
\(\Rightarrow\left(0,5+\frac{2}{3}-1\right).x=-4\)
\(\Rightarrow\frac{1}{6}.x=-4\)
\(\Rightarrow x=-4\div\frac{1}{6}\)
\(\Rightarrow x=-24\)
\(\text{Vậy }x=-24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : B = 23! + 19! + 15!
= 1.2.3.4..10.11...23 + 1.2.3.4...10.11..19 + 1.2.3.4..10.11..15
= 11(1.2.3...10.12.23 + 1.2.3.4..10.12...19 + 1.2.3.4....10.12...15) \(⋮\)11
b) Lại có B = 23! + 19! + 15!
= 1.2.3.4..10.11...23 + 1.2.3.4...10.11..19 + 1.2.3.4..10.11..15
= 10.11(1.2.3.4..9.12...23 + 1.2.3.4...9.12...19 + 1.2.3.4...9.12...15)
= 110(1.2.3.4..9.12...23 + 1.2.3.4...9.12...19 + 1.2.3.4...9.12...15) \(⋮\)110
BÀI GIẢI:
a) Ta có: B = 23! + 19! + 15!
= 1.2.3. ... .10 .11. ... 23 + 1.2.3. ... .10.11. ... .19 + 1.2.3. ... .10.11. ... .15
= 11 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15) \(⋮11\)
b) Ta có: B = 23! + 19! + 15!
= 1.2.3. ... .10 .11. ... 23 + 1.2.3. ... .10.11. ... .19 + 1.2.3. ... .10.11. ... .15
= 10.11 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15)
= 110 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15) \(⋮110\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)(99 hạng tử 1)
\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
Lại có \(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=200-\left(\frac{2}{1}+\frac{2}{2}+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\)
Khi đó
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+100\right)}=2\)(đpcm)
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\left(\text{có 99 hạng tử 1}\right)\)
\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
\(=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\text{Lại có : }200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)=200-\left(\frac{2}{1}+\frac{2}{2}+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\)
\(\text{Khi đó :}\)
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+100\right)}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có P = \(\frac{n^3-2n^2+3}{n-2}=\frac{n^2\left(n-2\right)+3}{n-2}=n^2+\frac{3}{n-2}\)
Để P \(\inℤ\Leftrightarrow3⋮n-2\Leftrightarrow n-2\inƯ\left(3\right)\Leftrightarrow n-2\in\left\{1;3;-1;-3\right\}\)
<=> \(n\in\left\{3;5;1;-1\right\}\)
Vậy \(n\in\left\{3;5;1;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+4+7+...+301\)
Tổng trên là tổng các số cách đều, số hạng sau hơn số hạng trước \(3\)đơn vị.
Tổng trên có số số hạng là:
\(\left(301\right)-1\div3+1=101\)(số hạng)
Giá trị của tổng trên là:
\(\left(301+1\right)\times101\div2=15251\)
Kim cha na