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\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
= \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{121}{120}\)
= \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{11^2}{10.12}\)
= \(\dfrac{2}{1}.\dfrac{2}{3}.\dfrac{3}{2}.\dfrac{3}{4}.\dfrac{4}{3}.\dfrac{4}{5}.....\dfrac{11}{10}.\dfrac{11}{12}\)
= \(\dfrac{2}{1}\left(\dfrac{2}{3}.\dfrac{3}{2}\right)\left(\dfrac{3}{4}.\dfrac{4}{3}\right)...\left(\dfrac{10}{11}.\dfrac{11}{10}\right).\dfrac{11}{12}\)
= \(2.\dfrac{11}{12}\)
= \(\dfrac{11}{6}\)
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{120}\right)\\ =\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\\ =\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{11^2}{10.12}\\ \)
\(=\frac{2.11}{1.12}=\frac{11}{6}\)
goi \(\frac{180\left(n-2\right)}{n}\)la so do cua mot da giac deu co n canh
goi \(\frac{180\left(m-2\right)}{m}\) la so do cua mot da giac deu co m canh
theo bai ra, ta co: \(\frac{\frac{180\left(n-2\right)}{n}}{\frac{180\left(m-2\right)}{m}}=\frac{5}{7}\)\(\Leftrightarrow\frac{180\left(n-2\right)}{n}.\frac{m}{180\left(m-2\right)}=\frac{5}{7}\)
\(\Leftrightarrow\frac{\left(n-2\right)m}{n\left(m-2\right)}=\frac{5}{7}\)\(\Leftrightarrow5mn-10n-7mn+14m=0\)
\(\Leftrightarrow\)-2mn-10n+14m=0\(\Leftrightarrow\)-2n(m+5)+14(m+5)=70
\(\Leftrightarrow\)(m+5)(7-n)=35=\(\pm1.\pm35=\pm5.\pm7\)
roi sau do ban tu giai nha.ket qua la n=6
a)x^2+5y^2+2x-4xy-10y+14
=x^2+2x-4xy+5y^2-10y+14
=x^2+2x(1-2y)+5y^2=10y+14
=x^2+2x(1-2y)+(1-2y)^2+5y^2-10xy-(1=2y)^2+14
=(x+1-2y)^2+5y^2-10y-(1-4y+4y^2)+14
=(x+1-2y)^2+5y^2-10y-1+4y-4y^2+14
=(x+1-2y)^2+y^2-6y+13
=(x+1-2y)^2+(y-3)^2+4
Vì....(đpcm)
b)5x^2+10y^2-6xy-4x-2y+3
=(x^2-6xy+9y^2)+(4x^2+1-4x)+(y^2-2y+1)+1
=(x-3y)^2+(2x-1)^2+(y-1)2+1
Vì....
(đpcm)
ta có: \(\dfrac{3x-4}{y+15}=\dfrac{1}{9}\)
theo đề bài, ta có phương trình:
\(\dfrac{3x-4}{12+15}=\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{7}{3}\)