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Bài 2:
a: Để E là số nguyên thì \(3n+5⋮n+7\)
\(\Leftrightarrow3n+21-16⋮n+7\)
\(\Leftrightarrow n+7\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
hay \(n\in\left\{-6;-8;-5;-9;-3;-11;1;-15;9;-23\right\}\)
b: Để F là số nguyên thì \(2n+9⋮n-5\)
\(\Leftrightarrow2n-10+19⋮n-5\)
\(\Leftrightarrow n-5\in\left\{1;-1;19;-19\right\}\)
hay \(n\in\left\{6;4;29;-14\right\}\)
bài 1
\(A=\left(\dfrac{3}{8}+\dfrac{1}{4}+\dfrac{5}{12}\right):\dfrac{7}{8}\)
\(A=\dfrac{9+6+10}{24}:\dfrac{7}{8}=\dfrac{25}{24}.\dfrac{8}{7}=\dfrac{25.1}{3.7}=\dfrac{25}{21}\)
\(B=\dfrac{1}{4}:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(B=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(B=\dfrac{1}{4}.2-\dfrac{3}{4}\)
\(B=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\)
\(M=-\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(M=-\dfrac{5}{7}\left(-\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\)
\(M=-\dfrac{5}{7}.\dfrac{7}{11}+\dfrac{12}{7}\)
\(M=-\dfrac{5}{11}+\dfrac{12}{7}=\dfrac{97}{77}\)
\(N=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(N=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3.4}{16}\)
\(N=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{13}{56}\)
Bài 2: chia 10n cho 5n-3 như bình thường ta được dư là 6
Để A có giá trị nguyên thì \(10n⋮5n-3\) Do đó 6 phai chia hết cho 3n+2
<= >5n-3\(\in u\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\\)
Lập bảng
| 5n-3= | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| n= | -0.6 | 0 | 0.2 | 0.4 | 0.8 | 1 | 1.2 | 1.8 |
\(1.a.\frac{x}{7}=\frac{6}{21}=\frac{6:3}{21:3}=\frac{2}{7}\Rightarrow x=2\\ b.\frac{-5}{y}=\frac{20}{28}=\frac{20:\left(-4\right)}{28:\left(-4\right)}=\frac{-5}{-7}\Rightarrow y=-7\)
\(2.a.\frac{a}{-b}=\frac{a\left(-1\right)}{-b\left(-1\right)}=\frac{-\left(a.1\right)}{-\left[-\left(b.1\right)\right]}=\frac{-a}{b}\\ b.\frac{-a}{-b}=\frac{-a\left(-1\right)}{-b\left(-1\right)}=\frac{-\left[-\left(a.1\right)\right]}{-\left[-\left(b.1\right)\right]}=\frac{a}{b}\)
\(3.\frac{3}{-4}=\frac{-3}{4}\\ \frac{-5}{-7}=\frac{5}{7}\\ \frac{2}{-9}=\frac{-2}{9}\\ \frac{-11}{-10}=\frac{11}{10}\)
\(4.\frac{3}{6}=\frac{2}{4}\\ \frac{6}{3}=\frac{4}{2}\\ \frac{2}{3}=\frac{4}{6}\\ \frac{3}{2}=\frac{6}{4}\)
Bài 1:
a, \(\frac{x}{7}\)=\(\frac{6}{21}\)⇒x.21=6.7⇒x.21=42⇒x=2
b,\(\frac{-5}{y}=\frac{20}{28}\)⇒-5.28= 20.y⇒-140=20.y⇒y =-7
Bài 2:
a, \(\frac{a}{-b}\)= \(\frac{a.\left(-1\right)}{-b.\left(-1\right)}\)=\(\frac{-a}{b}\)
b, \(\frac{-a}{-b}=\frac{-a.\left(-1\right)}{-b.\left(-1\right)}=\frac{a}{b}\)
Bài 3:
1,\(\frac{3}{-4}=\frac{-3}{4}\)
2,\(\frac{-5}{-7}=\frac{5}{7}\)
3,\(\frac{2}{-9}=\frac{-2}{9}\)
4,\(\frac{-11}{-10}=\frac{11}{10}\)
Bài 4 :
\(\frac{3}{6}=\frac{2}{4}\) ;
\(\frac{6}{3}=\frac{4}{2}\);
\(\frac{3}{2}=\frac{6}{4}\);
\(\frac{2}{3}=\frac{4}{6}\).
Bài 4:
a)Ta có: B= 23!+19!−15!
B=1.2.3.....11..23+1.2....11.19-1.2.....11.12.13.14.15
Vì 11 chia hết cho 11=>23! chia hết cho 11
19!chia hết cho 11
15! chia hết cho 11
=="
Câu 1:
A - B = \(1.2+2.3+...+98.99-1^2-...-98^2\)
\(=1\left(2-1\right)+2\left(3-2\right)+...+98\left(99-98\right)\)
\(=1+2+...+98\)
\(=99.49=4851\)
Câu 2:
a, \(A=5+5^2+...+5^{100}\)
\(5A=5^2+5^3+...+5^{101}\)
\(4A=5A-A=\left(5^2+5^3+...+5^{101}\right)-\left(5+5^2+5^{100}\right)\)
\(4A=5^{101}-5\Leftrightarrow4a+5=5^{101}\)
Lại có 4a+5 = 5^n => n = 101.
b,Gọi ước nguyên tố chung của tử và mẫu là d.
=> \(18n+3⋮d\) => \(7\left(18n+3\right)⋮d\)
=> \(24n+7⋮d\)=> \(6\left(24n+7\right)⋮d\)
=> \(6\left(24n+7\right)-7\left(18n+3\right)⋮d\)
\(\Leftrightarrow21⋮d\Rightarrow d=\left\{3;7\right\}\)
Với d = 3. \(21n+7⋮̸3\)
Với d = 7 => \(18n+3-21⋮d\Leftrightarrow18n-18⋮d\)
\(\Leftrightarrow18\left(n-1\right)⋮d\)\(\Rightarrow n-1⋮d\Leftrightarrow n=7k-1\)