Question

mọi ng giúp em với ạ;-;

Answer 1

\(1,\\ a,A=\left(3\sqrt{2}+2+7\right)\left(5\sqrt{2}-7\right)=\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)=50-49=1\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\sqrt{x}+1+\sqrt{x}-2=2\sqrt{x}-1\\ b,B< A\Leftrightarrow2\sqrt{x}-1< 1\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0< x< 1\)

\(3,\\ 1,\\ a,m=4\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\\ b,\text{PT có 2 nghiệm }\Leftrightarrow\Delta=m^2-4\left(m-1\right)\ge0\Leftrightarrow\left(m-2\right)^2\ge0\left(\text{luôn đúng}\right)\\ \text{Viét: }\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\\ \text{Mà }x_1=9x_2\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{m}{10}\Rightarrow x_1=\dfrac{9m}{10}\\x_1x_2=m\end{matrix}\right.\\ \Leftrightarrow\dfrac{9m}{10}\cdot\dfrac{m}{10}=m-1\\ \Leftrightarrow9m^2=100m-100\\ \Leftrightarrow\left[{}\begin{matrix}m=10\\m=\dfrac{10}{9}\end{matrix}\right.\)