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a)100-9:(372:3.y-1)-14=83
<=>100-9:(124y-1)-14=83
<=>86-9:(124y-1)=83
<=>9:(124y-1)=3
<=>124y-1=3
<=>124y=4
<=>y=\(\frac{1}{31}\approx0,3223\)
b)7260-120:24.y+924=528,3
<=>8184-5y=528,3
<=>5y=7655,7
<=>y=1531,14
c)2000-52:(615:3:y-15)-14=1984
<=>1986-52:(205:y-15)=1984
<=>52:(205:y-15)=2
<=>205:y-15=26
<=>205:y=41
<=>y=5
d)y+y.1/3+5/18=7/18
<=>4/3y=1/9
<=>y=1/12
e)13/15-(5/21+y)7/12=7/10
<=>7/12.(5/21+y)=1/6
<=>5/36+7/12y=1/6
<=>7/12y=1/36
<=>y=1/21\(\approx\)0,4762
h)y+2.y+3.y+...+10.y=49,5
<=>55y=49,5
<=>y=0,9
i)y.(1975/8.9+1885/9.10+1755/10.11+1579/11.12)=1/24
<=>7991,364899y=1/24
<=>y=33/6329161
tích nha mỏi tay quá
a)=\(\frac{1}{3}\)
b)=\(\frac{39}{70}\)
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
b. \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2003}\)= \(\frac{1}{2013}\)
c. \(5.\left(\frac{1}{14}+\frac{1}{84}+\frac{1}{204}+\frac{1}{374}\right)\)= 5. \(\frac{1}{11}\)= \(\frac{5}{11}\)
Mình biết 2 câu này thôi, thông cảm nhá...!!!
d) Ta có: 100-(1+1/2+1/3+1/4+...+1/100)
=1x100-(1+1/2+1/3+1/4+...+1/100)
=(1-1)+(1-1/2)+(1-1/3)+(1-1/4)+....+(1-1/100)
=1/2+2/3+3/4+...+99/100
\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)
\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)
....
Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được
Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}=1\)
=> 2x - 2021 = 1
=> 2x = 2022
=> x = 1011
Vậy x = 1011
a) \(\frac{12}{24}\)= \(\frac{50}{100}\); \(\frac{18}{15}\)= \(\frac{120}{100}\)
b) \(\frac{14}{35}\)= \(\frac{40}{100}\); \(\frac{9}{6}\)= \(\frac{15}{10}\); \(\frac{3}{4}\)= \(\frac{75}{100}\); \(\frac{625}{1000}\)
c) \(\frac{1}{2}\)< x < \(\frac{2}{3}\)
\(\Rightarrow\frac{6}{12}< x< \frac{8}{12}\)
\(\Rightarrow x=\frac{7}{12}\)
Vậy...
d)\(\frac{2005}{2006}\)và \(\frac{2005}{2007}\)
\(\Rightarrow\)\(\frac{2005}{2006}>\frac{2005}{2007}\)
Vì nếu so ánh hai phân số cùng tử thì soa sánh mẫu
+Phân số nào có mẫu lớn hơn thì bé hơn
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
\(1-\frac{y}{3}=\frac{1}{3}+\frac{2}{6}\)
\(\Rightarrow1-\frac{y}{3}=\frac{2}{6}+\frac{2}{6}\)
\(\Rightarrow1-\frac{y}{3}=\frac{4}{6}\)
\(\Rightarrow\frac{y}{3}=1-\frac{4}{6}\)
\(\Rightarrow\frac{y}{3}=\frac{2}{6}\)
\(\Rightarrow y=\frac{2.3}{6}\)
\(\Rightarrow y=1\)
a)1-\(\frac{y}{3}\)=2/3
y/3 =1-2/3
y/3 =1/3
\(\Rightarrow\)y=1
Vậy y=1
b)100/y-24/18=1/18
100/y =1/18+24/18
100/y=25/18
100/y=100/72
\(\Rightarrow\)y=72
Vậy y=72
C)14/27.y=36/12:18/14
14/27.y=3:9/7
14/27.y=7/3
y =7/3:14/27
y =9/2
k cho mk đi