Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+b}{c+d}\\ \Leftrightarrow\left(\dfrac{a}{b}\right)^{2011}=\left(\dfrac{c}{d}\right)^{2011}=\left(\dfrac{a+b}{c+d}\right)^{2011}\\ \Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\)

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ \Rightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\\ \dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\\ \Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)

ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)

\(\Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\) (ĐPCM)

Bài 2:

\(A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}-\dfrac{x\left(x^3-1\right)}{x^2+x+1}\)

\(=x\left(x+1\right)-x\left(x-1\right)\)

=x^2+x-x^2+x

=2x

a/ Ta có :

\(\left(x+y+t\right)-x^3-y^3-z^3=2011\)

\(\Leftrightarrow3\left(x+y\right)\left(y+t\right)\left(t+x\right)=2011\)

\(\Leftrightarrow\left(x+y\right)\left(y+t\right)\left(t+x\right)=\dfrac{2011}{3}\)

Thay vào D ta được :

\(D=\dfrac{2011}{\dfrac{2011}{3}}=3\)

Vậy.....

b/ Ta có :

\(H=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Leftrightarrow10899H=10899\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow10899H=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(\Leftrightarrow10899H=1+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+1+\dfrac{b}{c}+\dfrac{c}{b}+1\)

\(\Leftrightarrow10899H=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Áp dụng BĐT Cô - si cho các số dương ta có ;

\(+,\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

+, \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

+, \(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

Cộng vế với vế của các BĐT ta có :

\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\ge6\)

\(\Leftrightarrow10899H\ge9\)

\(\Leftrightarrow H\ge\dfrac{1}{2011}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=6033\)

Vậy..

b ) Do a ; b ; c dương \(\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\) dương

Áp dụng BĐT Cô - si cho 3 số dương , ta có :

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

Theo GT : \(a+b+c=18099\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{18099}=\dfrac{1}{2011}\)

\(\Rightarrow H\ge\dfrac{1}{2011}\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=18099\\a=b=c\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=6033\)

Vậy ...

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

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

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

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

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

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

\(1.\) 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 .

\(\RightarrowĐCPM\)

Đặt \(B=A\div C\)

\(C=2012+\dfrac{2011}{2}+...+\dfrac{1}{2012}=2012+\dfrac{2013-2}{2}+\dfrac{2013-3}{3}+...+\dfrac{2013-2012}{2012}\)

\(C=2012+\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2012}-1-1-...-1\)

\(C=2012+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)-2011\)

\(C=1+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)=\dfrac{2013}{2013}+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)

\(C=2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=2013.A\)

\(\Rightarrow B=\dfrac{A}{C}=\dfrac{1}{2013}\)

a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)

\(\Leftrightarrow\)\(2^x.15=480\)

\(\Rightarrow\)\(2^x=480:15\)

\(\Leftrightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5.