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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
Học tốt nha^^
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x=0
<=> x=0
Vậy x=0
+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)
\(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x.\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,6\right\}\)
+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)
\(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)
Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)
=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)
=> \(x^2-4x-2x+8-x-2=-2x\)
=> \(x^2-5x+6=0\)
=> \(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)
=> x = 3 .
Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)
b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)
Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)
=> \(x\left(x+12\right)=192\)
=> \(x^2+12x-192=0\)
=> \(x^2+2x.6+36-228=0\)
=> \(\left(x+6\right)^2=288\)
=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)
\(A=\frac{2^2-1^2}{\left(1.2\right)^2}+\frac{3^2-2^2}{\left(2.3\right)^2}+\frac{4^2-3^2}{\left(3.4\right)^2}+...+\frac{100^2-99^2}{\left(99.100\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)
\(A=1-\frac{1}{100^2}=\frac{9999}{10000}\)