Tìm x
\(\frac{x-199}{6}+\frac{x-199}{12}+\frac{x-199}{20}+\frac{x-199}{30}+\frac{x-199}{42}=\frac{5}{14}\)\(\frac{5}{14}\)
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Tìm x biết: \(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}\)
Ủa ko có vế phải thì mình làm bằng niềm tin à? :D
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Xem thêm câu trả lời
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
Đặt: \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{199}{1}\)là B
Cộng 1 vào mỗi phần số trừ phân số cuối cùng ta sẽ được:
B= \(\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)
=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}\)
=> B= \(200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\) => B= \(200\) X A
=> \(\frac{A}{B}\)\(=\frac{1}{200}\)
=> \(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
=>\(x-20\) =\(\frac{1}{2000}:\frac{1}{200}\)
=> \(x-20=\).......................... Bạn tự làm tiếp nhé, chúc bạn học tốt !!!^^\(\)
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CMR: \(14< \frac{2}{1}.\frac{4}{3}.\frac{6}{5}....\frac{200}{199}< 20\)
(x-20).\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)=\(\frac{1}{2000}\)
Giúp mk với
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)
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Cho A=\(\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}CMR:14< A< 20\)
tìm x,biết:a)frac{2}{left(x+2right)left(x+4right)}+frac{4}{left(x+4right)left(x+8right)}+frac{6}{left(x+8right)left(x+14right)}frac{x}{left(x+2right)left(x+14right)}b)frac{x+1}{10}+frac{x+1}{11}+frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}c)left(x+2right)^2frac{38}{25}+frac{9}{10}-frac{11}{15}+frac{13}{21}-frac{15}{28}+frac{17}{36}-...+frac{197}{4851}-frac{199}{4950}giúp tớ với,huhu
Đọc tiếp
tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
tìm x biết:\(\frac{x+1}{199}\)+\(\frac{x+2}{198}\)+\(\frac{x+3}{197}\)+\(\frac{x+4}{196}\)+\(\frac{x+5}{195}\)=-5
chuyển vế rồi tách 5 thành 5 số 1` rồi nhóm vào
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tìm x \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\left(x+1\right)}=\frac{199}{200}\)
= 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/x - 1/x + 1
= 1/2 - 1 /x + 1 =199/200
=1/x+1 = 1/2 - 199/200
1/ x + 1 = ....
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\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=-\frac{99}{200}\)
\(\Rightarrow\frac{-99}{-\left(x+1\right).99}=\frac{-99}{200}\)
\(\Rightarrow-99x+-99=200\)
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\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!
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