bằng 0.33333333333 

Ta có x-y-z=0=> x=y+z

=> A= x(yz-y^2-z^2) thay x=y+z vào A ta được

A= (y+z)(yz-y^2-z^2)=y^2z-y^3-z^2y+yz^2-zy^2-z^3=-y^3-z^3

mà B=y^3+z^3

=> A+B=-y^3-z^3+y^3+z^3=0(dpcm)

b)\(\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}}\)

\(=\frac{2.2014}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2014.2015}{2}}}\)

\(=\frac{2.2014}{2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)}\)

\(=\frac{2014}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}}\)

\(=\frac{2014}{1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}}\)

\(=\frac{2014}{1-\frac{1}{2015}}=\frac{2014}{\frac{2014}{2015}}=2015\)

a)\(\frac{\frac{105}{13.20}+\frac{105}{20.27}+...+\frac{105}{62.69}}{\frac{5}{9.13}+\frac{7}{9.25}+\frac{13}{19.25}+\frac{31}{19.69}}\)

\(=\frac{\frac{105}{7}.\left(\frac{7}{13.20}+\frac{7}{20.27}+...+\frac{7}{62.69}\right)}{\left(\frac{9}{9.13}-\frac{4}{9.13}\right)+\left(\frac{16}{9.25}-\frac{9}{9.25}\right)+\left(\frac{19}{19.25}-\frac{6}{19.25}\right)+\left(\frac{50}{19.69}-\frac{19}{19.69}\right)}\)

\(=\frac{\frac{105}{7}.\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+...+\frac{1}{62}-\frac{1}{69}\right)}{\left(\frac{1}{13}-\frac{1}{9}+\frac{1}{13}\right)+\left(\frac{1}{9}-\frac{1}{25}-\frac{1}{25}\right)+\left(\frac{1}{25}-\frac{1}{19}+\frac{1}{25}\right)+\left(\frac{1}{19}-\frac{1}{69}-\frac{1}{69}\right)}\)

\(=\frac{\frac{105}{7}.\left(\frac{1}{13}-\frac{1}{69}\right)}{\frac{1}{13}+\frac{1}{13}-\frac{1}{69}-\frac{1}{69}}=\frac{\frac{105}{7}\left(\frac{1}{13}-\frac{1}{69}\right)}{2\left(\frac{1}{13}-\frac{1}{69}\right)}=\frac{\frac{105}{7}}{2}=\frac{15}{2}\)

Câu hỏi của Nguyễn Thị Vân - Toán lớp 7 - Học toán với OnlineMath

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