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a, Với \(x\ge0;x\ne\frac{16}{9};4\)
\(P=\frac{2\sqrt{x}-4}{3\sqrt{x}-4}-\frac{4+2\sqrt{x}}{\sqrt{x}-2}+\frac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)
\(=\frac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-3x+\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{-\left(3\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{2-\sqrt{x}}\)
b, \(P\ge-\frac{3}{4}\Rightarrow\frac{\sqrt{x}+1}{2-\sqrt{x}}+\frac{3}{4}\ge0\Leftrightarrow\frac{4\sqrt{x}+4+6-3\sqrt{x}}{8-4\sqrt{x}}\ge0\Leftrightarrow\frac{\sqrt{x}+10}{8-4\sqrt{x}}\ge0\)
\(\Rightarrow2-\sqrt{x}\ge0\Leftrightarrow x\le4\)Kết hợp với đk vậy \(0\le x< 4\)
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
Ta có: \(Q=\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)
\(=\frac{3\left(-5\sqrt{x}+4\right)\left(2\sqrt{x}+3\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(6\sqrt{x}+4\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{3\left(-10x-7\sqrt{x}+12\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(18x-8\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{-30x-21\sqrt{x}+36+54x-24+29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{24x+8\sqrt{x}-16}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(3x+3\sqrt{x}-2\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(\sqrt{x}+1\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}+8}{6\sqrt{x}+9}\)
2) Để \(Q>\frac{8}{3}\) thì \(Q-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{8\sqrt{x}+8}{6\sqrt{x}+9}-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24}{3\left(6\sqrt{x}+9\right)}-\frac{8\left(6\sqrt{x}+9\right)}{3\left(6\sqrt{x}+9\right)}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24-48\sqrt{x}-72}{9\left(2\sqrt{x}+3\right)}>0\)
mà \(9\left(2\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(-24\sqrt{x}-48>0\)
\(\Leftrightarrow-24\left(\sqrt{x}+2\right)>0\)
\(\Leftrightarrow\sqrt{x}+2< 0\)(Vô lý)
Vậy: Không có giá trị nào của x thỏa mãn \(Q>\frac{8}{3}\)