Cho A=1.3.5...49
B=\(\dfrac{1.2.3...49.50}{2.4.6....48.50}\)
C=\(\dfrac{26}{2}.\dfrac{27}{2}...\dfrac{50}{2}\)
So sánh A,B,C
A=1.3.5.7..49
B=\(\frac{1.2.3.4...49.50}{2.4.6...48.50}\)
C=\(\frac{26}{2}.\frac{27}{2}....\frac{50}{2}\)
So sánh A,B,C
giải chi tiết giúp mik với! mik đang cần gấp:
Cho A= 1.3.5.7...49; B= \(\dfrac{26}{2}.\dfrac{27}{2}...\dfrac{50}{2}\); so sánh A và B
ba bạn nhanh nhất mik sẽ tick cho!
Cho A=1.3.5.7...49
B=\(\frac{1.2.3.4.5.6...48.49.50}{2.4.6...48.50}\)
C=\(\frac{26}{2}.\frac{27}{2}...\frac{50}{2}\)
So sánh A,B và C
CHO:
\(A=1.3.5.7.....48.49\)
\(B=\frac{1.2.3.4.....49.50}{2.4.6.....48.50}\)
\(C=\frac{26}{2}\).\(\frac{27}{2}\).......\(\frac{49}{2}\).\(\frac{50}{2}\)
SO SÁNH A, B VÀ C
giải ra giúp mk nhé!
ai nhanh mk tk~
ban len mang di , nam nay mk moi len lop 6
chuc ban hoc tot ^-^
hình như sai đề thì phải. Phần A đó, cuối cùng phải là 47.49 chứ
a) \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
b) \(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+...+\dfrac{4}{21.23.25}\)
c) \(\dfrac{3}{1.2}-\dfrac{5}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+...+\dfrac{39}{19.20}-\dfrac{41}{20.21}\)
d) \(\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{120}{121}\)
e) \(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
Các bạn không nhất thiết phải làm hết, làm cho nó dễ hiểu được thì càng tốt để mk vận dụng
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
So sánh A và B với \(\dfrac{1}{2}\) biết :
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+........+\dfrac{1}{49.50^2}\) và
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{50^2}\)
8 Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) ; b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100}{100!}\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
d) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.......+\dfrac{1}{3^{99}}\)
=> 3A=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^{98}}\)
=> 3A-A= 1-\(\dfrac{1}{3^{99}}\)
=> A=\(\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)
=> A<1/2
Vậy A<1/2
1.Chứng minh rằng:
a) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
b) Cho A = \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Chứng minh \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
2. Tìm a, b \(\in\) Q, biết
a - b = a.b = a : b
2, a-b=ab => a=ab+b => a=b(a+1)
thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1
Theo bài ra ta có: a:b=a-b
=> a+1=a-b
=>-b=1
=> b=-1
Thay b=-1 vào a-b=ab ta có : a-(-1)=-a
=> a +1=-a
=>a=-1/2
Vậy a=-1/2. b=-1
so sánh các số sau : \(a=\dfrac{35}{49};b=\sqrt{\dfrac{5^2}{7^2}};c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}};d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)