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Ta có \(\dfrac{a}{2009}\)=\(\dfrac{b}{2010}\)=\(\dfrac{c}{2011}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{c-a}{2011-2009}=\dfrac{c-a}{2}\left(1\right)\)
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{a-b}{2009-2010}=\dfrac{a-b}{-1}\)(2)\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{b-c}{2010-2011}=\dfrac{b-c}{-1}\left(3\right)\)
Từ (1),(2),(3) \(_{\Rightarrow}\)\(\dfrac{c-a}{2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\Rightarrow\dfrac{\left(a-c\right)^{ }2}{2^{ }2}=\dfrac{\left(a-b\right)}{-1}\times\dfrac{\left(b-c\right)}{-1}\)
\(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(a-b\right)\times\left(b-c\right)}{1}\Rightarrow4\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\)
\(\Rightarrow M=4\left(a-b\right).\left(a-c\right)-\left(c-a\right)^2=0\)
Vậy M = 0
đặt \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=k\) ta có:
\(\Rightarrow a=2009k\left(1\right)\\ \Rightarrow b=2010k\left(2\right)\\ \Rightarrow c=2011k\left(3\right)\)
thay 1;2;3 vào M ta có:
\(M=4\left(2009k-2010k\right)\left(2010k-2011k\right)-\left(2011k-2009k\right)^2\\ \Rightarrow M=4.\left(-k\right)\left(-k\right)-\left(2k\right)^2\\ \Rightarrow M=4k^2-\left(2k\right)^2\\ \Rightarrow M=\left(2k\right)^2-\left(2k\right)^2\\ \Rightarrow M=0\)Vậy M = 0
Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
$A=(|2x-4|+|2x-8|)+|2x-6|=(|2x-4|+|8-2x|)+|2x-6|$
$\geq |2x-4+8-2x|+|2x-6|$
$=4+|2x-6|\geq 4$
Vậy $A_{\min}=4$. Giá trị này đạt tại \(\left\{\begin{matrix}
(2x-4)(8-2x)\geq 0\\
2x-6=0\end{matrix}\right.\Leftrightarrow x=3\)
a,\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)=\(\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{19}.3^9+2^{18}.3^9.5}{6^9.2^{10}.\left(1+6\right)}\)=\(\dfrac{2^{18}.3^9.\left(2+5\right)}{6^9.2^{10}.7}\)=\(\dfrac{2^{18}.3^9}{6^9.2^{10}}=\dfrac{2^{10}.2^8.3^9}{2^9.3^9.2^{10}}=\dfrac{2^8}{2^8.2}=\dfrac{1}{2}\)
b, \(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{16}}{\dfrac{-23}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-29}{16}}{\dfrac{-29}{16}}=1\)
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)