\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)\(=1+\frac{\frac{2.3}{2}}{2}+\frac{\frac{3.4}{2}}{3}+...+\frac{\frac{199.200}{2}}{199}\)\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+...+\frac{199.200}{199.2}\)\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)\(=\frac{2+3+4+...+200}{2}\)\(=\frac{\frac{200.201}{2}}{2}\)\(=\frac{200.201}{2.2}\)\(=10050\)

trà my sai rồi,

nà ní ko có quy luật à 

\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)

\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(=50.\frac{9}{50}=9\)

Câu hỏi là j vậy bạn?

phân tích nhân tử đi -_-, làm ra dài nên lười viết, chỉ cho cách làm đáy

ra đc ko?

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...................

\(\frac{a-1}{a}:\left(\frac{a^2+1}{a^2+2a}-\frac{2}{a+2}\right)\)

\(=\frac{a-1}{a}:\left(\frac{a^2+1}{a\left(a+2\right)}-\frac{2}{a+2}\right)\)

\(=\frac{a-1}{a}:\left(\frac{a^2+1-2a}{a\left(a+2\right)}\right)\)

\(=\frac{a-1}{a}:\frac{\left(a-1\right)^2}{a\left(a+2\right)}\)

\(=\frac{a-1}{a}.\frac{a\left(a+2\right)}{\left(a-1\right)^2}\)

\(=\frac{a+2}{a-1}\)