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Khoảng cách có rồi thì bạn áp dụng công thức : \(\frac{a}{m.n}=\frac{1}{m}-\frac{1}{n}\)(với n-m=a) là làm được
S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{93.95}+\frac{3}{95.98}+\frac{4}{98.102}+\frac{5}{102.17}+\frac{2012}{107..2119}\)
S=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+\frac{1}{102}-\frac{1}{107}+\frac{1}{107}-\frac{1}{2119}\)
S=\(\frac{1}{5}-\frac{1}{2119}\)
S=\(\frac{2114}{10595}\)
4/3.5+4/5.7+4/7.9+4/9.11
=4.(1/3.5+1/5.7+1/7.9+1/9.11)
=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
4/3.5+4/5.7+4/7.9+4/9.11
=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11
=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11
=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
Ta có: \(3^x+3^{x+1}+3^{x+2}=351\)
\(\Rightarrow3^x.1+3^x.3+3^x.3^2^{ }=351\)
\(\Rightarrow3^x.1+3^x.3+3^x.9=351\)
\(\Rightarrow3^x.\left(1+3+9\right)=351\)
\(\Rightarrow3^x.13=351\)
\(\Rightarrow3^x=351:13=27\)
\(\Rightarrow x=3\)
3 k mk nha
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a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20
b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)
=>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008
Ta có: \(3\left|x^2-1\right|-6=\left|1-x^2\right|\)
\(\Leftrightarrow3\left|x^2-1\right|-\left|x^2-1\right|=6\)
\(\Leftrightarrow2\left|x^2-1\right|=6\)
\(\Leftrightarrow\left|x^2-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=3\\x^2-1=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=-2\end{cases}}\)
Vì \(x\ge0>-2\left(\forall x\right)\)
\(\Rightarrow x^2=4\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(x-\frac{5}{9}=\frac{-1}{3}\)
\(x=\frac{-1}{3}+\frac{5}{9}\)
\(x=\frac{2}{9}\)
x2 + xy + x + y = 2
x . x + x . y + x + y = 2
x . ( x + y ) + x + y = 2
x . ( x + y ) + ( x + y ) . 1 = 2
( x + y ) . ( x + 1 ) = 2
=> x + 1 thuoc U(2)
=> x + 1 thuoc { 1 ; 2 }
Lap bang :
x + 1 | 1 | 2 |
x + y | 2 | 1 |
x | 0 | 1 |
y | 2 | 1 |
Vay ( x ; y ) la : ( 0 ; 2 ) ; ( 1 ; 1 )
P/s tham khao nha
Đặt \(A=\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{x\left(x+2\right)}\)(sửa đề)
\(\Rightarrow A=\frac{1}{2}.3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{1}{2}-\frac{3}{2x+4}\)