1) đkxđ \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\y\ge0\end{matrix}\right.\)

Xét biểu thức \(P=x^3+y^3+7xy\left(x+y\right)\)

\(P=\left(x+y\right)^3+4xy\left(x+y\right)\)

\(P\ge4\sqrt{xy}\left(x+y\right)^2\)

Ta sẽ chứng minh \(4\sqrt{xy}\left(x+y\right)^2\ge8xy\sqrt{2\left(x^2+y^2\right)}\)  (*)

Thật vậy, (*)

\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{2xy\left(x^2+y^2\right)}\)

\(\Leftrightarrow\left(x+y\right)^4\ge8xy\left(x^2+y^2\right)\)

\(\Leftrightarrow x^4+y^4+6x^2y^2\ge4xy\left(x^2+y^2\right)\) (**)

Áp dụng BĐT Cô-si, ta được:

VT(**) \(=\left(x^2+y^2\right)^2+4x^2y^2\ge4xy\left(x^2+y^2\right)\)\(=\) VP(**)

Vậy (**) đúng \(\Rightarrowđpcm\). Do đó, để đẳng thức xảy ra thì \(x=y\). 

Thế vào pt đầu tiên, ta được \(\sqrt{2x-3}-\sqrt{x}=2x-6\)

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

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

 Rõ ràng với \(x\ge\dfrac{3}{2}\) thì \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}\le\dfrac{1}{\sqrt{\dfrac{2.3}{2}-3}+\sqrt{\dfrac{3}{2}}}< 2\) nên ta chỉ xét TH \(x=3\Rightarrow y=3\) (nhận)

Vậy hệ pt đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(3;3\right)\)

ĐKXĐ \(3x^2-5x+1\ge0;x^2-2\ge0;x^2-x-1\ge0\)

Ta có : \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3.\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\sqrt{3x^2-5x+1}-\sqrt{3\left(x^2-x-1\right)}=\sqrt{x^2-2}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\dfrac{3x^2-5x+1-3.\left(x^2-x-1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}=\dfrac{x^2-2-x^2+3x-4}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

\(\Leftrightarrow\dfrac{-2x+4}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}=\dfrac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}+\dfrac{2}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}=0\left(∗\right)\end{matrix}\right.\)

Xét phương trình (*) ta có VT > 0 \(\forall x\) mà VP = 0

nên (*) vô nghiệm

Vậy x = 2 là nghiệm phương trình 

Điều kiện: \(\left\{\begin{matrix}x\left(x-1\right)\ge0\\x\left(x+2\right)\ge0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\le0\text{∨}x\ge1\\x\le-2\text{∨}x\ge0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=0\\x\ge1\end{matrix}\right.\)

Với \(x=0\) thì \(\left(\text{*}\right)\Leftrightarrow0=0\Rightarrow x=0\)là 1 nghiệm của \(\left(\text{*}\right)\).

Với \(x\ge1\) thì \(\left(\text{*}\right)\Leftrightarrow\sqrt{x}\left(\sqrt{x-1}+\sqrt{x+2}\right)=2\sqrt{x^2}\Leftrightarrow\sqrt{x-1}+\sqrt{x+2}=2\sqrt{x}\)

\(\Leftrightarrow x-1+x+2+2\sqrt{\left(x-1\right)\left(x+2\right)}=4x\Leftrightarrow\sqrt{\left(x-1\right)\left(x+2\right)}=x-\frac{1}{2}\)\(\Leftrightarrow\left\{\begin{matrix}x\ge\frac{1}{2}\\x^2+x-2=x^2-x+\frac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ge\frac{1}{2}\\x=\frac{9}{8}\end{matrix}\right.\Leftrightarrow x=\frac{9}{8}\left(N\right)\)

Vậy phương trình có hai nghiệm là \(x=0\text{∨}x=\frac{9}{8}.\)

\(Xét-mẫu-của-biểu-thức:\left(đk:x\ge1\right).ta-có:x-\sqrt{2\left(x^2+5\right)}=\frac{-\left(x^2+10\right)}{x+\sqrt{2\left(x^2+5\right)}}< 0\\ .\)Vậy nó luôn <0 với đk x>=1
\(Xét-tử:đặt-nó-bằng-A=\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)=2\sqrt{x-1}\left(2x-1\right)-\left(x-1\right)\left(x+4\right)\\ =\sqrt{x-1}\left(2\left(2x-1\right)-\sqrt{x-1\left(x+4\right)}\right)\ge0.\\ \)\(=>\left(2\left(2x-1\right)-\sqrt{\left(x-1\right)}\left(x+4\right)\right)\ge0< =>\frac{\left(5-x\right)\left(x-2\right)^2}{2\left(2x-1\right)+\left(x-1\right)\left(x+4\right)}\ge0< =>x\le5\) Vậy . \(1\le x\le5\)
 

Thank you ^^^

a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)

=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)

c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)

=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)

=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)

=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)

=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)

=>\(x=\dfrac{1}{4}\Omega+k\Omega\)