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Ta có
\(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}\)
\(=\frac{n^2+n-2}{\left(n+1\right)n}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Áp dụng vào bài toán ta có
\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+x}\right)=\frac{672}{2017}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+1\right)}=\frac{672}{2017}\)
\(\Leftrightarrow\frac{1}{3}.\frac{x+2}{x}=\frac{672}{2017}\)
\(\Leftrightarrow2017x+4034=2016x\)
- \(\Leftrightarrow x=-4034\)
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/3
<=>1/x-1/x+1+1/x+1-1/x+2+1/x+2-1/x+3+1/x+3-1/x+4=1/3
<=>1/x-1/x+4=1/3
<=>x+4/x(x+4)-x/x(x+4) ( quy dong mau ) =1/3
<=>4/x(x+4)=1/3
<=> 4.3=x(x+4) ( nhan cheo )
<=> x(x+4)=12
<=> x^2+4x-12=0
<=>x^2-2x+6x-12=0
<=>x(x-2) + 6(x-2) =0
<=> (x-2)(x+6)=0
<=> x-2 =0 hoac x +6=0
<=>x=2 hoac x= -6
Vay x thuoc ( 2,-6 )
K mk nha !!
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x\text{+}2\right)}\text{+}\frac{1}{\left(x\text{+}2\right)\left(x\text{+}3\right)}+\frac{1}{\left(x\text{+}3\right)\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x\text{+}1}\text{+}\frac{1}{x\text{+}1}-\frac{1}{x\text{+}2}\text{+}.....\text{+}\frac{1}{x\text{+}3}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\)\(\frac{1}{x}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\frac{x\text{+}4}{x\left(x\text{+}4\right)}-\frac{x}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{4}{12}\)
\(\Rightarrow x\left(x\text{+}4\right)=12\)
mà x và x+4 cách nhau 4 đơn vị \(\Rightarrow x=2\)và x+4\(=\)6
Vậy \(x=2\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
Ta có
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)
\(\Rightarrow1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)
\(=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Áp dụng vào bài toán ta được
\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+x}\right)=\frac{672}{2017}\)
\(\Leftrightarrow\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+1\right)}=\frac{672}{2017}\)
\(\Leftrightarrow\frac{1}{3}.\frac{\left(x+2\right)}{x}=\frac{672}{2017}\)
\(\Leftrightarrow2016x=2017\left(x+2\right)\)
Đề có thể bị sai rồi bạn
\(\Leftrightarrow x=\)