Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành: 

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)

\(\Rightarrow x=2018;y=2019;z=2020\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)

\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)

\(x=2018,y=2019,z=2020\)

ĐK: \(x\ge0\)

Với \(x\ge0\Rightarrow\sqrt{\left(x+1\right)^3}-\sqrt{x}>0\)nên bpt \(\Leftrightarrow\sqrt{x\left(x+2\right)}\ge\sqrt{\left(x+1\right)^3}-\sqrt{x}\)

\(\Leftrightarrow x^2+2x\ge x^3+3x^2+4x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\)

\(\Leftrightarrow x^3+2x^2+2x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\le0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2+x+1-2\sqrt{x\left(x+1\right)}\right]\le0\)

\(\Leftrightarrow x^2+x+1-2\sqrt{x\left(x+1\right)}\le0\)

\(\Leftrightarrow\left(\sqrt{x\left(x+1\right)}-1\right)^2\le0\Leftrightarrow\sqrt{x\left(x+1\right)}-1=0\)

\(\Leftrightarrow x=\frac{-1\pm\sqrt{5}}{2}.dox\ge0\Rightarrow x=\frac{-1+\sqrt{5}}{2}\)

\(a.b.c=1\)

\(\Leftrightarrow a=1,b=1,c=1\)

\(a+bc=a+b.c=1+1.1=\left(1+1\right)\left(1+1\right)\)

Vậy \(a+b.c=\left(a+b\right)\left(a+c\right)\).

(đúng không?)

Ta có : \(abc=1\Leftrightarrow a=1;b=1;c=1\)(1)

Ta có : \(a+bc=\left(a+b\right)\left(a+c\right)\Leftrightarrow VT=a^2+ac+ab+bc\)(2)

Thay (1) vào (2) ta được : 

\(1+1=1^2+1+1+1\Leftrightarrow2\ne4\)

GiáTrị của Biểu thức là:

\(\left(-3\right)\sqrt{2}\sqrt{11}\sqrt{g}\sqrt{t}+3\sqrt{2}\sqrt{11}+2\sqrt{3^3}\sqrt{5}\)

Ta có:\(x=\sqrt[3]{15+3\sqrt{22}}+\sqrt[3]{15-3\sqrt{22}}\Rightarrow x^3=\left(\sqrt[3]{15+3\sqrt{22}}\right)^3+\left(\sqrt[3]{15-3\sqrt{22}}\right)^3+3\sqrt[3]{\left(15+3\sqrt{22}\right)\left(15-3\sqrt{22}\right)}\left(\sqrt[3]{15+3\sqrt{22}}+\sqrt[3]{15-3\sqrt{22}}\right)\)\(\Rightarrow x^3=15+3\sqrt{22}+15-3\sqrt{22}+3\sqrt[3]{27}x\Rightarrow x^3=30+9x\Rightarrow x^3-9x+1981==2011\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\Rightarrow\left(a+c\right)\left(b+c\right)=c^2\)

Vì \(a,b>0\)mà \(\frac{1}{c}=-\left(\frac{1}{a}+\frac{1}{b}\right)< 0\)nên \(c< 0\Rightarrow\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)

\(\Rightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\Rightarrow\left(a+c\right)+2\sqrt{\left(a+c\right)\left(b+c\right)}+\left(b+c\right)=a+b\)

\(\Rightarrow\left(\sqrt{a+c}+\sqrt{b+c}\right)^2=a+b\)---> 2 vế đều dương nên ta lấy căn 2 vế:

\(\sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\)