tìm x biết:\(\frac{x+1}{199}\)+\(\frac{x+2}{198}\)+\(\frac{x+3}{197}\)+\(\frac{x+4}{196}\)+\(\frac{x+5}{195}\)=-5
Bai 1: Tim x biet :
\(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)
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 =
tim x:
cau 1:\(\frac{x+1}{199}\)+\(\frac{x+2}{198}\)+\(\frac{x+3}{197}\)+\(\frac{x+4}{196}\)+\(\frac{x+5}{195}\)=(-5)
cau 2:\(\frac{1}{21}\)+\(\frac{1}{28}\)+\(\frac{1}{36}\)+....+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2}{9}\)
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
Câu 2: => 2/42 + 2/56 + 2/72 + ... + 2/x(x+1) = 2/9
=> 2/6*7+2/7*8+...+2/x(X+1) = 2/9
\(\frac{x+1}{199}+\frac{x+2}{198}+\frac{x+3}{197}+\frac{x+4}{196}+\frac{x+5}{195}=-5\)
\(\left(\frac{x+1}{199}+1\right)+\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)+\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)=0\)
\(\frac{x+200}{199}+\frac{x+200}{198}+\frac{x+200}{197}+\frac{x+200}{196}+\frac{x+200}{195}=0\)
\(\left(x+200\right)\left(\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+\frac{1}{196}+\frac{1}{195}=0\right)\)
\(x+200=0\left(Vì\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+\frac{1}{196}+\frac{1}{195}\right)\)
\(\Rightarrow x=-200\)
Bài 1: Tìm x biết:
a. \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
b. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+..+\frac{4}{41.45}=\frac{29}{45}\)
c. \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+2\right)}:2\)
d. (x-20) . \(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2000}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
Bài 2:
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\right)\in Nsao\)
Bài 3:
a)\(\frac{3}{x}+\frac{y}{3}=\frac{5}{6}\)
b) \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
c) \(\frac{x}{2}+\frac{y}{3}=\frac{x+y}{2+3}\)
d) \(\frac{x-1}{9}+\frac{1}{3}=\frac{1}{y+2}\)
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:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
C1. Tìm \(\frac {a}b\), biết rằng :
a, \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
b, \(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
c, \(C=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)
C2. Tìm x :
1. \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1991}{1993}\)
2. \(\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{11\times14}+...+\frac{1}{x\times(x+3)}=\frac{101}{1540}\)
Tìm số tự nhiên x,y biết :x-3=y*(x+2)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
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 !!!^^\(\)
tìm x
x+1/199 + x+2/198 + x+3/197 + x+4/196 + x+5/195 = -5
1/21 +1/28 +1/36 +........+2/ x(x+1)=2/9
(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
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)
tìm x
a)\(\frac{x-5}{195}\)+ \(\frac{x-7}{193}\)= \(\frac{x-2}{198}\)+ \(\frac{x-3}{197}\)
b) \(\frac{x+4}{96}\)+ \(\frac{x+5}{95}\)= \(\frac{x+7}{93}\)+ \(\frac{x+9}{91}\)
\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)
\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)
\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)
Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)
=> x + 399 = 0
=> x = -399