tìm x biết:\(\frac{x+1}{199}\)+\(\frac{x+2}{198}\)+\(\frac{x+3}{197}\)+\(\frac{x+4}{196}\)+\(\frac{x+5}{195}\)=-5
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Ta có : \(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)
=> \(\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)=\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)\)
=> \(\frac{x+2+198}{198}+\frac{x+3+197}{197}=\frac{x+4+196}{196}+\frac{x+5+195}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
=> \(\left(x+200\right)\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)=0\)
Ta có : \(\frac{1}{198}+\frac{1}{197}\ne\frac{1}{196}+\frac{1}{195}\) => \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\ne0\)
=> x + 200 = 0
=> x = -200
<=> (\(\frac{x+2}{198}\)+1) +(\(\frac{x+3}{197}\)+1) =(\(\frac{x+4}{196}\)+1) +(\(\frac{x+5}{195}\)+1)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
<=> \(\left(x+200\right)\cdot\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)\)=0
Vì \(\frac{1}{195}>\frac{1}{196}>\frac{1}{197}>\frac{1}{198}\)
<=> \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\) khác 0
<=> \(x+200=0\)
<=> x =
ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200
tung từng vế một thôi
bạn nhác quá éo chịu suy nghĩ
bài này dễ vl
Bài 1:
a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
c, ghi lại đề
d, ghi lại đề
Bài 2:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
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 !!!^^\(\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)
\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)
\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)
\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)
Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)
=> x + 399 = 0
=> x = -399
chuyển vế rồi tách 5 thành 5 số 1` rồi nhóm vào