câu 1 đề sai hay vô nghiệm ko bt

câu 2: pt thứ 2 thiếu

nếu chưa ai làm chiều học về mk sẽ làm

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

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

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

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

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

aaa là \(\sqrt{x+3}\) cháu gõ lộn

Ta có: \(\frac{1}{2\sqrt3-2}-\frac{3}{2\left(\sqrt3+1\right)}\)

\(=\frac{1}{2\left(\sqrt3-1\right)}-\frac{3}{2\left(\sqrt3+1\right)}\)

\(=\frac{\sqrt3+1-3\left(\sqrt3-1\right)}{2\left(\sqrt3-1\right)\left(\sqrt3+1\right)}=\frac{\sqrt3+1-3\sqrt3+3}{2\cdot2}=\frac{-2\sqrt3+4}{4}=\frac{4-2\sqrt3}{4}\)

=>\(x=\sqrt{\frac{4-2\sqrt3}{4}}=\sqrt{\frac{\left(\sqrt3-1\right)^2}{2^2}}=\frac{\sqrt3-1}{2}\)

\(4\left(x+1\right)\cdot x^{2013}-2\cdot x^{2012}+2x+1\)

\(=2x^{2012}\left\lbrack2x\left(x+1\right)-1\right\rbrack+2x+1\)

\(=2x^{2012}\cdot\left\lbrack2\cdot\frac{\sqrt3-1}{2}\cdot\left(\frac{\sqrt3-1}{2}+1\right)-1\right\rbrack+2\cdot\frac{\sqrt3-1}{2}+1\)

\(=2x^{2012}\cdot\left\lbrack\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}{2}-1\right\rbrack+\sqrt3-1+1=\sqrt3\)

\(2x^2+3x\)

\(=2\left(\frac{\sqrt3-1}{2}\right)^2+3\cdot\frac{\sqrt3-1}{2}=\frac{2\cdot\left(4-2\sqrt3\right)}{4}+\frac{3\left(\sqrt3-1\right)}{2}\)

\(=\frac{4-2\sqrt3+3\sqrt3-3}{2}=\frac{\sqrt3+1}{2}\)

Ta có: \(A=\frac{4\left(x+1\right)\cdot x^{2013}-2\cdot x^{2012}+2x+1}{2x^2+3x}\)

\(=\sqrt3:\frac{\sqrt3+1}{2}=\sqrt3\cdot\frac{2}{\sqrt3+1}=\sqrt3\left(\sqrt3-1\right)=3-\sqrt3\)

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....