Tìm x, biết:
(12+22+32+...+492)x(2-x)=\(\frac{-11}{5}\)
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-12(x-5) + 21 - 7 = 5
=> -12(x-5) = 5+7-21 = -9
=> x-5= -9:(-12)= 3/4
=> x = 3/4+5=23/5
a) \(3^2.\frac{1}{243}.81^3.\frac{1}{33}\)
\(=\frac{3^2}{243}.\frac{81^3}{33}\)
\(=\frac{3^2}{3^5}.\frac{3^{12}}{3.11}\)
\(=\frac{1}{3^3}.\frac{3^{11}}{11}\)
\(=\frac{3^8}{11}.\)
b) \(42^5:\left(2^3.\frac{1}{16}\right)\)
\(=42^5:\frac{2^3}{16}\)
\(=42^5:\frac{2^3}{2^4}\)
\(=42^5:\frac{1}{2}\)
\(=42^5.2\)
\(=21^5.2^5.2\)
\(=21^5.2^6.\)
c) \(12^5.\frac{12^2}{9^3.4^5}\)
\(=\frac{12^7}{9^3.4^5}\)
\(=\frac{3^7.4^7}{3^6.4^5}\)
\(=3.4^2\)
\(=48.\)
d) \(\frac{2^5+2^6+2^7}{2^7+2^8+2^9}\)
\(=\frac{2^5\left(1+2+2^2\right)}{2^7\left(1+2+2^2\right)}\)
\(=\frac{2^5}{2^7}\)
\(=\frac{1}{4}.\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b-c}{b+c-d}\right)^3\)
Mà \(\left(\frac{a}{b}\right)^3=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)
=>\(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{a}=\frac{\left(b+c-d\right)^3}{d}\) (đpcm)
ta có S = 1.2.3 + 2.3.4 + ..... + 198.199.200
4s = 1.2.3.4 + 2.3.4.(5-1)+....+198.199.200.(201-197)
= 1.2.3.4+2.3.4.5-1.2.3.4+...+198.199.200.201-197.198.199.200
= 198.199.200.201
=> s = \(\frac{198.199.200.201}{4}=198.199.50.201\)