bạn kl nốt đoạn cuối nha. lười

\(ĐK:x\ge-2\)

cái này là lớn nhất hay nhỏ nhất v bạn

\(\Leftrightarrow2\sqrt{(2x^2+16x+8)\left(x^2+1\right)}=x^2-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3\ge0\\4\left(2x^2+16x+8\right)\left(x^2+1\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{3}\le x\le\sqrt{3}\\4\left(2x^4+16x^3+10x^2+16x+8\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow7x^4+64x^3+46x^2+64x+23=0\)

\(\sqrt{2x^2+16x+18}+\sqrt{x^2+1}=2x+4\left(1\right)\)

\(ĐK:x\in R\)

\(pt\left(1\right)\Leftrightarrow2x^2+16x+18+x^2+1+2\sqrt[]{(2x^2+16x+18)\left(x^2+1\right)}=4x^2+16x+16\)

\(\Leftrightarrow3+2\sqrt{(2x^2+16x+18)\left(x^2+1\right)}=x^2\)

\(a)A\ge\dfrac{x-\sqrt{x}-3}{\sqrt[]{x}}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}}\ge\dfrac{x-\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow\sqrt{x}-4\ge x-\sqrt{x}-3\)

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

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

\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)

\(b)ĐKXĐ:x>0;x\ne4\)

\(B=\dfrac{x+2\sqrt{x}-10}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

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

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

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

\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\left(đpcm\right)\)

\(c)\dfrac{A}{B}=\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\Rightarrow\left|\dfrac{A}{B}\right|=\dfrac{\left|\sqrt{x}-4\right|}{\sqrt{x}+3}\left(vì\sqrt{x}+3>0\right)\)

Xét các TH:

\(TH1:\sqrt{x}-4< 0\Leftrightarrow\sqrt{x}< 4\Leftrightarrow x< 16\left(1\right)\)

\(\Rightarrow\left|\dfrac{A}{B}\right|=\dfrac{4-\sqrt{x}}{\sqrt{x}+3}\)

\(\left|\dfrac{A}{B}\right|>\dfrac{A}{B}\Leftrightarrow\dfrac{4-\sqrt{x}}{\sqrt{x}+3}>\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\)

\(\Leftrightarrow4-\sqrt{x}>\sqrt{x}-4\Leftrightarrow2\sqrt{x}< 8\Leftrightarrow\sqrt{x}< 4\)

\(\Leftrightarrow x< 16\left(2\right)\)

Từ (1)(2) suy ra x<16 suy ra x lớn nhất bằng 15 

\(TH2:\sqrt{x}-4\ge0.\) Giai tương tự TH1 suy ra loại

cái đoạn\(-xy\left(x+1\right)\)đổi x+1 thành y+1 nha mik đánh nhầm