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1) a.Ta có \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)
Vì \(3\inℤ\Rightarrow\frac{21}{n-4}\inℤ\Rightarrow21⋮n-4\Rightarrow n-4\inƯ\left(21\right)\)
=> \(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)
=> \(n\in\left\{5;3;8;1;11;-3;25;-17\right\}\)
b) Ta có B = \(\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)
Vì \(3\inℤ\Rightarrow\frac{8}{2n-1}\inℤ\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)(1)
lại có với mọi n nguyên => 2n \(⋮\)2 => 2n - 1 không chia hết cho 2 (2)
Kết hợp (1) ; (2) => \(2n-1\in\left\{1;-1\right\}\Rightarrow n\in\left\{1;0\right\}\)
2) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
=> \(\frac{20+xy}{4x}=\frac{1}{8}\)
=> 4x = 8(20 + xy)
=> x = 2(20 + xy)
=> x = 40 + 2xy
=> x - 2xy = 40
=> x(1 - 2y) = 40
Nhận thấy : với mọi y nguyên => 1 - 2y là số không chia hết cho 2 (1)
mà x(1 - 2y) = 40
=> 1 - 2y \(\inƯ\left(40\right)\)(2)
Kết hợp (1) (2) => \(1-2y\in\left\{1;5;-1;-5\right\}\)
Nếu 1 - 2y = 1 => x = 40
=> y = 0 ; x = 40
Nếu 1 - 2y = 5 => x = 8
=> y = -2 ; x = 8
Nếu 1 - 2y = -1 => x = -40
=> y = 1 ; y = - 40
Nếu 1 - 2y = -5 => x = -8
=> y = 3 ; x =-8
Vậy các cặp (x;y) thỏa mãn là : (40 ; 0) ; (8; - 2) ; (-40 ; 1) ; (-8 ; 3)
4) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{-4}{3}}=\frac{-\frac{5}{60}}{\frac{2}{5}}=-\frac{5}{60}:\frac{2}{5}=-\frac{5}{24}\)
b) \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)
\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
c) \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{121}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{120}{121}\)
\(-A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot10\cdot12}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot11\cdot11}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot...\cdot11\right)\left(2\cdot3\cdot4\cdot...\cdot11\right)}\)
\(-A=\frac{1\cdot12}{11\cdot2}=\frac{6}{11}\)
\(A=-\frac{6}{11}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(B=1-\frac{1}{38}=\frac{37}{38}\)
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
a)Ta có :
\(\frac{4}{5}=\frac{8}{10}\)
1,1=\(\frac{11}{10}\)
Vì \(\frac{8}{10}< \frac{11}{10}\)
»\(\frac{4}{5}< 1,1\)
vậy \(\frac{4}{5}< 1,1\)