CM:\(3< 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 6\)
A=\(\dfrac{2}{3}\)+\(\dfrac{14}{15}\)+\(\dfrac{34}{35}\)+\(\dfrac{62}{63}\)+\(\dfrac{98}{99}\)+\(\dfrac{142}{143}\)+\(\dfrac{194}{195}\)
Và B=5+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^3}\)+\(^{\dfrac{1}{4^4}}\)+\(\dfrac{1}{5^5}\)+\(\dfrac{1}{6^6}\)+\(\dfrac{1}{7^7}\).So sánh A và B
1,\(\dfrac{5\left(x-1\right)+2}{6}\)-\(\dfrac{7x-1}{4x}\)=\(\dfrac{2\left(2x+1\right)}{7}\)-5
2,\(\dfrac{3\left(x-3\right)}{4}\)+\(\dfrac{4x-10,5}{10}\)=\(\dfrac{3 \left(x+1\right)}{5}\)+6
3,\(\dfrac{2\left(3x+1\right)+1}{4}\)-5=\(\dfrac{2\left(3x-1\right)}{5}\)-\(\dfrac{3x+2}{10}\)
Diễn giải ra cho em với ạ!Em cảm ơn
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
chứng tỏ B = \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 6\)
Ta có :
\(B=1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{63}\)
Ta thấy :
\(1=1\)
\(\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{1+1}+\dfrac{1}{1+2}< \dfrac{2}{1+1}=\dfrac{2}{2}=1\)
\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}=\dfrac{1}{3+1}+\dfrac{1}{3+2}+\dfrac{1}{3+3}+\dfrac{1}{3+4}< \dfrac{4}{3+1}=\dfrac{4}{4}=1\)
\(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}=\dfrac{1}{7+1}+\dfrac{1}{7+2}+....+\dfrac{1}{7+8}< \dfrac{8}{7+1}=\dfrac{8}{8}=1\)
\(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}=\dfrac{1}{15+1}+\dfrac{1}{15+2}+...+\dfrac{1}{15+16}< \dfrac{16}{15+1}=\dfrac{16}{16}=1\)
\(\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{63}=\dfrac{1}{31+1}+\dfrac{1}{31+2}+...+\dfrac{1}{31+32}< \dfrac{32}{31+1}=\dfrac{32}{32}=1\)
\(\Rightarrow B< 1+1+....+1\) (\(6\) số 1)
\(\Rightarrow B>6\rightarrowđpcm\)
tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính nhanh
A = \(\dfrac{1}{2}\) + \(^{\dfrac{1}{3}}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{63}\)
Mình đang cần gấp , mong mn giúp mình với ạ
quy đòng r tính nha ra \(\dfrac{199}{33}\)
Cho biểu thức S=\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{63}\)
Chứng minh rằng 3<S<6
Để chứng minh 3<S<6, ta cần tính giá trị của biểu thức S và thấy xem nó có nằm trong khoảng (3, 6) hay không.
Đầu tiên, ta tính tổng S bằng cách đặt S bên cạnh tổng harmonic thứ 63, rồi trừ đi tổng harmonic thứ 62:
S = 1/1 + 1/2 + 1/3 + ... + 1/63 S - 1/2 = 1/2 + 1/3 + ... + 1/63
Lặp lại phương pháp trên đối với S - 1/2, ta có:
S - 1/2 - 1/3 = 1/3 + ... + 1/63
Cứ lặp lại phương pháp trên đến khi ta được:
S - 1/2 - 1/3 - ... - 1/62 = 1/63
Tổng quát lại, ta có:
S - 1/2 - 1/3 - ... - 1/62 - 1/63 = 0
Từ đây suy ra:
3/2 < 1/2 + 1/3 + ... + 1/62 + 1/63 < 1 + 1/2 + 1/3 + ... + 1/62 < 6
Vì vậy, ta có:
3 < S < 6
Vậy, ta đã chứng minh được rằng 3<S<6.
CM: \(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}+\dfrac{3}{\sqrt{6}-3}\right).\dfrac{5}{9\sqrt{6}+4}=\dfrac{1}{2}\)
\(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}-\dfrac{3}{3-\sqrt{6}}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\left(\dfrac{2+2\sqrt{6}}{5}+\dfrac{6+3\sqrt{6}}{2}-3-\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{4+4\sqrt{6}+30+15\sqrt{6}-30-10\sqrt{6}}{10}\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{1}{2}\)
Chứng tỏ rằng:
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{62}+\dfrac{1}{63}+\dfrac{1}{64}>4\)