Tính: B = \(\frac{4}{3}\times\frac{9}{8}\times...\times\frac{100}{99}-\frac{9}{11}\)
tìm x biết
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times....\times\frac{100}{99}+\frac{9}{110}\)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
Cho biểu thức A= \(\frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\frac{8}{7}\times\frac{10}{9}\times...\times\frac{100}{99}\)Chứng minh rằng 12<A<13
\(\frac{1}{3}\times\frac{2}{5}\times\frac{3}{7}\times\frac{4}{9}\times\frac{5}{11}\times\frac{6}{15}\times\frac{7}{15}\times\frac{8}{15}\times\frac{9}{19}\times\frac{10}{21}\times\frac{11}{32}\times\frac{12}{25}\times\left\{\frac{126}{252}-\frac{2}{4}\right\}\)
Để nhân các phân số này, ta chỉ cần nhân tử số với nhau và mẫu số với nhau:
\[
\frac{1}{3} \times \frac{2}{5} \times \frac{3}{7} \times \frac{4}{9} \times \frac{5}{11} \times \frac{6}{15} \times \frac{7}{15} \times \frac{8}{15} \times \frac{9}{19} \times \frac{10}{21} \times \frac{11}{32} \times \frac{12}{25} \times \left( \frac{126}{252} - 4 \right)
\]
Sau đó, ta thực hiện các phép tính:
1. Nhân tử số:
\[1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12 \times 126 = 997920\]
2. Nhân mẫu số:
\[3 \times 5 \times 7 \times 9 \times 11 \times 15 \times 15 \times 15 \times 19 \times 21 \times 32 \times 25 \times 252 = 7621237680\]
Kết quả là:
\[\frac{997920}{7621237680}\]
Bây giờ, ta có thể rút gọn phân số này bằng cách chia tử số và mẫu số cho 160:
\[ \frac{997920}{7621237680} = \frac{997920 ÷ 160}{7621237680 ÷ 160} = \frac{6237}{47695230} \]
1.Tính nhanh
a,\(\frac{1}{1\times4}+\frac{1}{4\times7}+............+\frac{1}{97\times100}\)
b,\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...........\times\frac{99}{100}\)
c,\(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...........\times\frac{99}{100}\)
d,\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\left(\frac{1}{4}+1\right)\times............\times\left(\frac{1}{99}+1\right)\)
e,\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times..........\times\left(1-\frac{1}{100}\right)\)
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
a) \(\frac{4}{15}\times\frac{7}{9}+\frac{4}{15}\times\frac{2}{9}\)
b)\(\frac{13}{19}\times\frac{23}{11}-\frac{13}{19}\times\frac{8}{11}-\frac{13}{19}\times\frac{4}{11}\)
Giúp mình với nhé!
\(a.\) \(\frac{4}{15}.\frac{7}{9}+\frac{4}{15}.\frac{2}{9}\)
\(=\frac{4}{15}\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{4}{15}.\frac{9}{9}\)
\(=\frac{4}{15}.1\)
\(=\frac{4}{15}\)
\(b.\) \(\frac{13}{19}.\frac{23}{11}-\frac{13}{19}.\frac{8}{11}-\frac{13}{19}.\frac{4}{11}\)
\(=\frac{13}{19}\left(\frac{23}{11}-\frac{8}{11}-\frac{4}{11}\right)\)
\(=\frac{13}{19}.\frac{11}{11}\)
\(=\frac{13}{19}.1\)
\(=\frac{13}{19}\)
a)4/15 x(7/9+2/9)=4/15x1=4/15
b)13/19x(23/11-8/11-4/11)13/19x1=13/19
A) \(\frac{4}{15}x\frac{7}{9}+\frac{4}{15}x\frac{2}{9}\)
\(=\frac{28}{135}+\frac{8}{135}\)
\(=\frac{36}{135}=\frac{4}{15}\)
B) \(\frac{13}{19}x\frac{23}{11}-\frac{13}{19}x\frac{8}{11}-\frac{13}{19}x\frac{4}{11}\)
\(=\frac{299}{209}-\frac{104}{209}-\frac{52}{209}\)
\(=\frac{143}{209}\)
1) Rút gọn biểu thức M:
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{2}{11}}\)
2) Tính nhanh:
\(A=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times....\times\left(1-\frac{1}{100}\right)\)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
\(A=\left(1\frac{1}{6}\times\frac{6}{7}\times6:\frac{3}{5}\right):\left(4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{10}\right)\)
\(B=1\frac{13}{15}\times25\%\times3+\left(\frac{8}{15}-\frac{79}{60}\right):1\frac{23}{4}\)
\(C=\frac{123}{4567}\times\frac{1}{8}+\frac{123}{4567}\times\frac{1}{2}-\frac{123}{4567}\times\frac{13}{8}\)
\(D=\frac{10\frac{1}{3}\times\left(24\frac{1}{2}-15\frac{6}{7}\right)-\frac{12}{11}\times\left(\frac{10}{3}-1,75\right)}{\left(\frac{5}{9}-0,25\right)\times\frac{60}{11}+194\frac{8}{99}}\)
Tính giá trị biểu thức:
\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+\frac{4}{11}\times\frac{4}{15}+...+\frac{4}{95}\times\frac{4}{99}\)
tính các tích sau
\(a=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{9999}{10000}\)
\(b=\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(c=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(d=\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99\times100}\right)\)
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)