\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)
\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)
\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)
Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)
=> x + 399 = 0
=> x = -399
Tìm x biết: \(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}\)
Tính tỉ số A và B , biết rằng :
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
Tìm x, biết:
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
D=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b) \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)
A=\(\frac{1}{199}-\frac{1}{199\cdot198}-\frac{1}{198-197}-\frac{1}{197-196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
Bai 1: Tim x biet :
\(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)
BT1: Tìm chữ số tận cùng của \(2015^{2014}-2014^{2015}\)
BT2: Cho đa thức: \(f\left(x\right)=a\times x+b\) biết \(f\left(1\right)=1;f\left(2\right)=4\)Tìm a;b
BT3: Cho \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\left(b\ne0\right)\)Tìm c
BT4: Tìm x biết: \(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)
BT5 Tính \(S=1\times2+2\times3+3\times4+4\times5+...+49\times50\)
Bạn nào giải xong trước mình kích cho nhé(nhớ giải hết bài của mình nha!)!!!
tim x
a,\(\left(x-20\right)+\left(x-19\right)+\left(x-18\right)+....+200+201=201\)
b, \(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+....+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)
