Rút gọn:
P=(12+22+32+42+........+20152):(22+42+62+........+40302)
giúp mik vs,mai mik phải nộp bài này cho cô rồi,Thanks
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SSH:(20152-12):10+1=2015
(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152
-10+(-10)+(-10)+...+(-10)+20152
-10x(2015-1):2+20152=12
=> C=12
A=1/2^2+1/3^2+...+1/10^2
=>A<1-1/2+1/2-1/3+...+1/9-1/10=1-1/10<1
\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)
=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)
=> \(\frac{15}{32}-x=\frac{3}{4}-2\)
=> \(\frac{15}{32}-x=-\frac{5}{4}\)
=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)
b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)
=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{10x-x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)
=> x = 2
Lời giải:
\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)
\(=4A=4.385=1540\)
\(P=\left(1^2+2^2+...............+2015^2\right):\left(2^2+4^2+........+4030^2\right)\)
\(P=\left(1^2+2^2+............+2015^2\right):\left[\left(1.2\right)^2+\left(2.2\right)^2+.............+\left(2.2015\right)^2\right]\)
\(P=\left(1^2+2^2+........+2015^2\right):\left(1^2.2^2+2^2.2^2+...............+2015^2.2^2\right)\)
\(P=\left(1^2+2^2+......+2015^2\right):2^2.\left(1^2+2^2+.........+2015^2\right)\)
\(P=\left(1^2+2^2+........+2015^2\right).\frac{1}{2^2.\left(1^2+2^2+..............+2015^2\right)}\)
\(P=\frac{1^2+2^2+...............+2015^2}{2^2.\left(1^2+2^2+............+2015^2\right)}=\frac{1}{2^2}=\frac{1}{4}\)
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