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a, Với x >= 0 ; x khác 4
\(=\frac{x-3\sqrt{x}+2-\left(x+4\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-3\sqrt{x}-3-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-7\sqrt{x}-6-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)
b, \(Q+1>0\Leftrightarrow\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}>0\Leftrightarrow\frac{-8}{\sqrt{x}-2}>0\)
\(\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\Rightarrow0\le x< 4\)
c, \(\frac{-\left(\sqrt{x}+6\right)}{\sqrt{x}-2}=\frac{-\left(\sqrt{x}-2+8\right)}{\sqrt{x}-2}=-1-\frac{8}{\sqrt{x}-2}\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
| \(\sqrt{x}-2\) | -1 | 1 | -2 | 2 | -4 | 4 | -8 | 8 |
| x | 1 | 9 | 0 | 16 | loại | 36 | loại | 100 |
\(T=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(\Rightarrow T=\frac{x-1}{\sqrt{x}}\left(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x-1}\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\)
\(\Rightarrow T=\frac{x-1}{\sqrt{x}}.\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)
\(\Rightarrow T=\frac{x-1}{\sqrt{x}}.\frac{2x+2}{x-1}\)
\(\Rightarrow T=\frac{2x+2}{\sqrt{x}}\)
\(T=8\Leftrightarrow\frac{2x+2}{\sqrt{x}}=8\)
\(\Leftrightarrow x+1=4\sqrt{x}\)
\(\Leftrightarrow x^2+2x+1=8x\)
\(\Leftrightarrow x^2-6x+1=0\)
\(\Delta=\left(-6\right)^2-4.1.1=36-4=32,\sqrt{\Delta}=\sqrt{32}\)
Vậy pt có 2 nghiệm phân biệt x1; x2
\(x_1=\frac{6+\sqrt{32}}{2}=3+\sqrt{8}\);\(x_2=\frac{6-\sqrt{32}}{2}=3-\sqrt{8}\)
N=\(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right).\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)
ĐKXĐ \(\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\3-x\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\x\ne3\\x\ge0\end{cases}}\)
\(=\left[\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right].\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right).\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\frac{x-2x+3}{3-x}=\frac{3-x}{3-x}=1\)
câu 2 ra |a-b| nha bn mik đăng rồi nhưng bị lỗi nên nó ko hiện lên
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((