Tên của mày là Tôm

bài này cũng khó đấy!

Câu c,

Đặt a/2009=b/2010=c/2011=k

=>a=2009.k (1)

   b=2010.k (2) 

   c=2011.k (3)

Thay (1),(2),(3) vào biểu thức:

M=4.(a-b)(b-c)-(c-a)^2 ,ta được:

M=4.(2009.k-2010.k)(2010.k-2011.k)-(2011.k-2009.k)^2

M=4.(-k).(-k)-(2k)^2

M=4.k^2-4.k^2

M=0.

Ta có:\(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\dfrac{x^{2010}}{a^2}=\dfrac{y^{2010}}{b^2}=\dfrac{z^{2010}}{c^2}=\dfrac{t^{2010}}{d^2}\)

\(\Rightarrow\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}=\dfrac{x^{2010}}{a^2}\)

\(\Rightarrow\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}=0\)

\(\Leftrightarrow3\cdot\dfrac{y^{2010}}{b^2}=0\)

\(\Leftrightarrow y^{2010}=0\)

\(\Leftrightarrow y=0\)

CMTT\(\Rightarrow x=z=t=0\)

\(\Rightarrow T=0\)

\(1.\) Giả sử : \(a\ge b\ge c\Rightarrow a+b\ge a+c\ge b+c\)

Ta có : \(\dfrac{c}{a+b}\le\dfrac{c}{b+c};\dfrac{b}{a+c}\le\dfrac{b}{b+c};\dfrac{a}{b+c}=\dfrac{a}{b+c}\)

\(\Rightarrow\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\le\dfrac{b+c}{b+c}+\dfrac{a}{b+c}=1+\dfrac{a}{b+c}< 1+1=2\left(đpcm\right)\)

\(2.\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)

\(\Leftrightarrow x^2y+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2=0\)

\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)y\left(x+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)

+) Với : \(x=-y\) , ta có :

Đpcm \(\Leftrightarrow-\dfrac{1}{y^{2011}}+\dfrac{1}{y^{2011}}+\dfrac{1}{z^{2011}}=\dfrac{1}{-y^{2011}+y^{2011}+z^{2011}}\)

\(\Leftrightarrow\dfrac{1}{z^{2011}}=\dfrac{1}{z^{2011}}\left(luôn-đúng\right)\)

Tương tự với 2 TH còn lại .

\(\RightarrowĐCPM\)

a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)

\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)

\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)

\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)

\(=5\dfrac{4}{23}.23\)

\(=\dfrac{119}{23}.23\)

\(=\dfrac{119}{23}\)

b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)

\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{3}{2}\)

\(=\dfrac{-2}{3}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{9}{6}\)

\(=\dfrac{5}{6}\)

c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)

\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)

\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)

\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)

\(=1-1\)

\(=0\)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

(đợi đã, mình chưa tìm được hướng làm...)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\left(\dfrac{-60}{132}+\dfrac{42}{132}-\dfrac{-16}{132}-\dfrac{15}{132}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\dfrac{-17}{132}:\left(\dfrac{381}{22}-\dfrac{865}{22}\right)\)

\(=\dfrac{-17}{132}:\left(-22\right)\)

\(=\dfrac{-17}{132}.\dfrac{1}{-22}\)

\(=\dfrac{-17}{-2904}=\dfrac{17}{2904}\)