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Hello bạn, mk cx tên Mai nek.
\(\frac{2}{5}.\left(x-1\right)+1=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}\left(x+1\right)=\frac{3}{5}-1\)
\(\Rightarrow\frac{2}{5}\left(x+1\right)=-\frac{2}{5}\)
\(\Rightarrow x+1=-\frac{2}{5}:\frac{2}{5}\)
\(\Rightarrow x+1=-1\)
\(\Rightarrow x=-1-1\)
\(\Rightarrow x=-2\)
\(\left(\frac{2}{7}\times x+1\right)\times\left(3-\frac{1}{2}\times x\right)=0\)
\(TH1:\frac{2}{7}\times x+1=0\)
\(\frac{2}{7}\times x=-1\)
\(x=-\frac{2}{7}\)
\(TH2:3-\frac{1}{2}\times x=0\)
\(\frac{1}{2}\times x=3\)
\(x=\frac{3}{2}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{7}\right\}\)
(x+1)+(x+2)+...+(x+10)=2015
(x+x+..+x)+(1+2+...+10)=2015
10x+55=2015
10x=2015-55
10x=1950
x=1950/10
x=195
Số số hạng là :
(x + 10) - (x + 1) + 1 = 10 (số hạng)
Tổng trên là :
[(x + 10) . (x + 1)] . 10 : 2 = (2x + 11) . 10 : 2 = 2015
=> (2x + 11) . 10 = 2015 . 2 = 4030
=> 2x + 11 = 4030 : 10 = 403
=> 2x = 403 - 11 = 392
=> x = 392 : 2 = 196
Vậy x = 196
a/ (x+1)+(x+2)+(x+3)+...+(x+9)+(x+10)
= x+1 + x+2 + x+3 + ... + x+9 + x+10
= (x+x+x +...+ x+x ) + 1+2+3+...+9+10
= 10.x +55
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1007}{2015}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1007}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4030}\)
=>x+1=4030
=>x=4029
vậy x=4029
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
tự làm tiếp nhé mk ăn cơm đã
<=> 10x-(1+2+...+10)=-2015
<=> 10x-55=-2015
<=>10x=-1960
<=>x=-196