câu 1 đề sai hay vô nghiệm ko bt

câu 2: pt thứ 2 thiếu

\(x=\dfrac{\sqrt{\sqrt{5}-2}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+1\right)}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(x=\dfrac{1+\sqrt{5}-2}{\sqrt{3-\sqrt{5}}}-\left(\sqrt{2}-1\right)=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{6-2\sqrt{5}}}-\left(\sqrt{2}-1\right)\)

\(x=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{\left(\sqrt{5}-1\right)^2}}-\sqrt{2}+1=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{2}+1=1\)

\(\Rightarrow x^{2012}+2x^{2013}+3x^{2014}=1^{2012}+2.1^{2013}+3.1^{2014}=6\)

b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)

\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)

\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)

c/ ĐKXĐ: \(x\ge3\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>2013\\y>2014\\z>2015\end{matrix}\right.\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2013}-1}{x-2013}+\frac{1}{4}-\frac{\sqrt{y-2014}-1}{y-2014}+\frac{1}{4}-\frac{\sqrt{z-2015}-1}{z-2015}=0\)

\(\Leftrightarrow\frac{x-2013-4\sqrt{x-2013}+4}{4\left(x-2013\right)}+\frac{y-2014-4\sqrt{y-2014}+4}{4\left(y-2014\right)}+\frac{z-2015-4\sqrt{z-2015}+4}{4\left(z-2015\right)}=0\)

\(\Leftrightarrow\left(\frac{\sqrt{x-2013}-2}{2\sqrt{x-2013}}\right)^2+\left(\frac{\sqrt{y-2014}-2}{2\sqrt{y-2014}}\right)^2+\left(\frac{\sqrt{z-2015}-2}{2\sqrt{z-2015}}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2013}-2=0\\\sqrt{y-2014}-2=0\\\sqrt{z-2015}-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)

b/ Trừ vế cho vế 2 pt ta được:

\(x^3-y^3=2\left(y-x\right)\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy+2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2\right]=0\)

\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

Thay vào pt đầu:

\(x^3+1=2x\Leftrightarrow x^3-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow...\)

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....