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\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)\(=1+\frac{\frac{2.3}{2}}{2}+\frac{\frac{3.4}{2}}{3}+...+\frac{\frac{199.200}{2}}{199}\)\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+...+\frac{199.200}{199.2}\)\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)\(=\frac{2+3+4+...+200}{2}\)\(=\frac{\frac{200.201}{2}}{2}\)\(=\frac{200.201}{2.2}\)\(=10050\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
1)\(ĐKXĐ:x\ne2;x\ne-2\)
đầu bài..
.\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=x\left(3x-2\right)+1\)
\(\Leftrightarrow-6x^2-11x+2+9x^2-14x-8=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\Leftrightarrow x=-\frac{7}{23}\)(nhận)
Vậy...........
2).......\(ĐKXĐ:x\ne2;x\ne7\)
\(\Rightarrow\left(x+1\right)\left(x-7\right)=x-2\)
\(\Leftrightarrow x^2-6x-7=x-2\)
\(\Leftrightarrow x^2-7x-5=0\)..............
Vậy........
3)ĐKXĐ:\(x\ne1\)
.........\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\frac{7}{19}\)(nhận)
4)ĐKXĐ:\(x\ne-1\)
.........\(\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\)
\(\Leftrightarrow15x=5\Leftrightarrow x=\frac{1}{3}\)(nhận)
Vậy...................
\(\frac{a-1}{a}:\left(\frac{a^2+1}{a^2+2a}-\frac{2}{a+2}\right)\)
\(=\frac{a-1}{a}:\left(\frac{a^2+1}{a\left(a+2\right)}-\frac{2}{a+2}\right)\)
\(=\frac{a-1}{a}:\left(\frac{a^2+1-2a}{a\left(a+2\right)}\right)\)
\(=\frac{a-1}{a}:\frac{\left(a-1\right)^2}{a\left(a+2\right)}\)
\(=\frac{a-1}{a}.\frac{a\left(a+2\right)}{\left(a-1\right)^2}\)
\(=\frac{a+2}{a-1}\)
\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=\left(\frac{1}{200}-1\right)\left(\frac{1}{200}+1\right)\left(\frac{1}{199}-1\right)\left(\frac{1}{99}-1\right)...\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)
\(=\frac{-199}{200}.\frac{201}{200}.\frac{-198}{199}.\frac{200}{199}...\frac{-100}{101}.\frac{102}{101}\)
\(=\left(-\frac{199}{200}.\frac{-198}{199}...\frac{-100}{101}\right)\left(\frac{201}{200}.\frac{200}{199}...\frac{102}{101}\right)\)
\(=\frac{100}{200}.\frac{201}{101}=\frac{201}{202}\)