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 \(\left(x;y\right)=\left(3;3\right)\)

a. Đề bài sai, phương trình không giải được

b.

ĐKXĐ: \(x\ge-\dfrac{2}{3}\)

\(\left(2x+10\right)\left(\dfrac{1-\left(3+2x\right)}{1+\sqrt{3+2x}}\right)^2=4\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{\left(2x+10\right)4.\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}=4\left(x+1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)^2=0\Rightarrow x=-1\\2x+10=\left(1+\sqrt{3+2x}\right)^2\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow2x+10=2x+4+2\sqrt{2x+3}\)

\(\Leftrightarrow\sqrt{2x+3}=3\)

\(\Leftrightarrow x=3\)

\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)

\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)

c, ĐK: \(x-2>0\Rightarrow x>2\)

\(2log_5\left(x-2\right)=log_59\\ \Leftrightarrow log_5\left(x-2\right)^2=log_59\\ \Leftrightarrow\left(x-2\right)^2=3^2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.

d, ĐK: \(x-1>0\Leftrightarrow x>1\)

\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)

\(\sqrt{9\left(x-2\right)^2}+2x=\sqrt{x-1}\left(ĐK:x\ge1\right)\)

<=> \(\left|9\left(x-2\right)\right|+2x=\sqrt{x-1}\)

<=> \(\left[{}\begin{matrix}-9\left(x-2\right)+2x=\sqrt{x-1}\\9\left(x-2\right)+2x=\sqrt{x-1}\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2,40.....\left(TM\right)\\1,71....\left(TM\right)\end{matrix}\right.\)

b.

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

\(\Leftrightarrow x^4+2x^2-4+x\sqrt{2x^2+4x}=0\)

Đặt \(x\sqrt{2x^2+4x}=t\Rightarrow t^2=x^2\left(2x^2+4x\right)=2\left(x^4+2x^2\right)\)

Pt trở thành:

\(\dfrac{t^2}{2}-4+t=0\)

\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4x}=2\left(x>0\right)\\x\sqrt{2x^2+4x}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\left(x>0\right)\\x^4+2x^2-8=0\left(x< 0\right)\end{matrix}\right.\)

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

a.

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\Rightarrow\dfrac{2x^2+9}{x^2}=\dfrac{1}{t^2}\)

Pt trở thành:

\(\dfrac{1}{t^2}+2t=3\)

\(\Rightarrow2t^3-3t^2+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vô-nghiệm\right)\\4x^2=2x^2+9\left(x< 0\right)\end{matrix}\right.\)

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

Kiểm tra lại vế trái đề bài câu b