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

a,-a

b,số nguyên dương , 0

c,0 

Vì a;b;c > 0 nên \(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)

BĐT Cosi :

\(9a+\dfrac{1}{a}\ge2.\sqrt{9a.\dfrac{1}{a}}=2.3=6\\ 9b+\dfrac{1}{b}\ge6\\ 9c+\dfrac{1}{c}\ge6\\ \Rightarrow\left(9a+\dfrac{1}{a}\right)+\left(9b+\dfrac{1}{b}\right)+\left(9c+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Dấu "=" xảy ra khi a=b=c=1/3

a.

TH1 : x<1

\(\Rightarrow-x+1+3-x=5\\ \Rightarrow-2x+4=5\\ \Rightarrow-2x=1\\ \Rightarrow x=-\dfrac{1}{2}\left(tm\right)\)

TH2: \(1\le x< 3\)

\(\Rightarrow x-1+3-x=5\\ \Rightarrow0x=3\\ \Rightarrow x\in\varnothing\)

TH3: \(x\ge3\)

\(\Rightarrow x-1+x-3=5\\ \Rightarrow2x-4=5\\ \Rightarrow2x=9\\ \Rightarrow x=\dfrac{9}{2}\left(tm\right)\)

\(x^4+4\\ =\left(x^4+4x^2+4\right)-4x^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ =\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)

tui giải sai hả , s im re z

\(\left(x^4+2x^2+1\right)+2.\dfrac{x}{2}\left(x^2+1\right)+\dfrac{x^2}{4}-\dfrac{25}{4}x^2=0\\ \Rightarrow\left(x^2+1\right)^2+2.\dfrac{x}{2}\left(x^2+1\right)+\left(\dfrac{x}{2}\right)^2-\left(\dfrac{5}{2}x\right)^2=0\\ \Rightarrow\left(x^2+1+\dfrac{x}{2}\right)^2-\left(\dfrac{5}{2}x\right)^2=0\\ \Rightarrow\left(x^2-2x+1\right)\left(x^2+3x+1\right)=0\)

Tự giải típ :V

\(\dfrac{3\left(x-11\right)}{4}=\dfrac{3\left(x+1\right)}{5}\\ \Rightarrow\dfrac{3x-33}{4}=\dfrac{3x+3}{5}=\dfrac{3x-33-3x-3}{4-5}\\ =\dfrac{-36}{-1}=36\\ \Rightarrow3\left(x-11\right)=144\\ \Rightarrow x-11=48\\ \Rightarrow x=59\)

\(M=\left(\dfrac{y}{x\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3\left(x-3\right)}{3.x\left(x+3\right)}-\dfrac{x^2}{3.x.\left(x+3\right)}\right)\\ =\left(\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3.x.\left(x+3\right)}\right)\\ =\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3.x.\left(x+3\right)}{3x-9-x^2}\\ =\dfrac{\left(y+x^2-3x\right).3}{\left(x-3\right)\left(3x-9-x^2\right)}\\ \)