a) A = \(\dfrac{8}{9}\) . \(\dfrac{15}{16}\) . \(\dfrac{24}{25}\). ... .\(\dfrac{2499}{2500}\). Tính
b) Tìm các số nguyên n để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên.
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Tính \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.............................\dfrac{2499}{2500}\)
A=2.4/3^2 . 3.5/4^2 . 4.6/5^2 ............ . 49.51/50^2
A=2/3-51/50
A=17/25.
Chúc bạn hok tốt.
Bài này cũng dễ ý mà, vô cùng đơn giản.........
Giải:
Ta có: \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}.\)
\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{49.51}{50^2}.\)
\(=\dfrac{\left(2.3.4.....49\right)\left(4.5.6.....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}.\)
\(=\dfrac{2.51}{3.50}.\)
\(=\dfrac{17}{25}.\)
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A= 3^2-1/3.3 . 4^2-1/4.4 . 5^2-1/5.5 . ... 50^2-1/50.50 A= (3+1).(3-1).(4+1).(4-1).(5+1).(5-1). ... (50+1).(50-1) / 3.3.4.4.5.5. ... . 50.50 A=4.2.5.3.6.4. ... 51.49 / 3.3.4.4.5.5....50.50 A=(4.5.6. ... .51).(2.3.4. ... 49)/(3.4.5.... .50).(3.4.5.. ... 50) A= 51.2/3.50 A=17/25
Ta có:
\(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}......\dfrac{2499}{2500}\)
= \(\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}......\dfrac{49.51}{50.50}\)
= \(\dfrac{2.4.3.5.4.6......49.51}{3.3.4.4.5.5......50.50}\)
= \(\dfrac{\left(2.3.4....49\right)\left(4.5.6....51\right)}{\left(3.4.5....50\right)\left(3.4.5....50\right)}\)
= \(\dfrac{2}{50}.\dfrac{51}{3}\) = \(\dfrac{17}{25}\)
a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\).
Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)
\(A=\dfrac{n+4}{n+1}\) làm tương tự.
b) Dễ thấy các số ở mẫu có thể viết dưới dạng:
\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)
\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)
\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)
...
\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)
Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{3}{4}+1-\dfrac{8}{9}+1-\dfrac{15}{16}+1-\dfrac{24}{25}...+1-\dfrac{2499}{2500}\)
\(=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+...+\dfrac{1}{2500}\right)\)
Lại có: \(49-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+...+\dfrac{1}{50.50}\right)< 49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)
Mà \(49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)
\(=49-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(=49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4942}{102}\) \(\notin Z\)
Vậy B không phải là số nguyên
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)
\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)
\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)
\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)
\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)
\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)
Giải:
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)
\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)
\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)
\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)
Vậy \(A=\dfrac{30}{31}.\)
Bài 2:
a: -2*(-27)=54
6*9=54
=>Hai phân số này bằng nhau
b: -1/-5=1/5=5/25<>4/25
Bài 3:
a: =>16/x=-4/5
=>x=-20
b: =>(x+7)/15=-2/3
=>x+7=-10
=>x=-17
a) Ta có \(A=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\dfrac{4\cdot6}{5\cdot5}\cdot...\cdot\dfrac{49\cdot51}{50\cdot50}\)
\(=\dfrac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\cdot\dfrac{4\cdot5\cdot6\cdot...\cdot51}{3\cdot4\cdot5\cdot...\cdot50}\)
= \(\dfrac{2}{50}\cdot17=\dfrac{17}{25}\)
b) Vì n nguyên nên 3n - 1 nguyên
Để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên thì 12 ⋮ ( 3n - 1 ) hay ( 3n - 1 ) ϵ Ư( 12 )
Ư( 12 ) = { \(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\) }
Lập bảng giá trị
Vì n nguyên nên n ϵ { 0; 1; -1 }
Vậy n ϵ { 0; 1; -1 } để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên