\(a,B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\\ B=x-\sqrt{x}+1-\sqrt{x}=\left(\sqrt{x}-1\right)^2\)

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

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

\(a,m=-\dfrac{3}{2}\Leftrightarrow x^2-2\cdot\dfrac{1}{2}\cdot x-\dfrac{3}{2}+1=0\\ \Leftrightarrow x^2-x-\dfrac{1}{2}=0\\ \Leftrightarrow2x^2-2x-1=0\\ \Delta'=1+2=3\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{3}}{2}\\x=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\\ b,\text{PT có }n_o\Leftrightarrow\Delta'=\left(m+2\right)^2-\left(m+1\right)\ge0\\ \Leftrightarrow m^2+3m+3\ge0\\ \Leftrightarrow\left(m+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge0\left(\text{luôn đúng}\right)\)

Vậy PT có nghiệm với mọi m

\(c,\text{Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)\\x_1x_2=m+1\end{matrix}\right.\\ x_1\left(1-2x_2\right)+x_2\left(1-2x_1\right)=m^3\\ \Leftrightarrow\left(x_1+x_2\right)-4x_1x_2=m^3\\ \Leftrightarrow2\left(m+2\right)-4\left(m+1\right)=m^3\\ \Leftrightarrow m^3+2m=0\\ \Leftrightarrow m\left(m^2+2\right)=0\Leftrightarrow m=0\)

Áp dụng định lí PTG: \(AC=\sqrt{BC^2-AB^2}=16\left(cm\right)\)

Vậy \(S_{ABC}=\dfrac{1}{2}AB\cdot AC=\dfrac{1}{2}\cdot12\cdot16=96\left(cm^2\right)\)

Câu 1:

\(a,=12\sqrt{2}+8\sqrt{2}-3\sqrt{2}=17\sqrt{2}\\ b,=\dfrac{3-\sqrt{7}+3+\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}=\dfrac{6}{2}=3\\ c,M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-3\right]\left(\sqrt{x}+3\right)\\ M=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)=x-9\)

Câu 2:

\(ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}+3\sqrt{x+5}-2\sqrt{x+5}=12\\ \Leftrightarrow\sqrt{x+5}=\dfrac{12}{3}=4\Leftrightarrow x+5=16\\ \Leftrightarrow x=11\left(tm\right)\)

Câu 3:

\(b,\text{PT hoành độ giao điểm: }2x-4=x-1\Leftrightarrow x=3\Leftrightarrow y=2\Leftrightarrow A\left(3;2\right)\\ \text{Vậy }A\left(3;2\right)\text{ là tọa độ giao điểm}\\ c,\text{PT hoành độ giao điểm: }mx-3=2x-4\\ \text{Mà }x=5\Leftrightarrow5m-3=6\Leftrightarrow m=\dfrac{9}{5}\)

Đặt \((n_{MgO};n_{CuO})=(a;b)\)

\(MgO+H_2SO_4\to MgSO_4+H_2O\\ CuO+H_2SO_4\to CuSO_4+H_2O\\ \Rightarrow n_{MgSO_4}:n_{CuSO_4}=1:1=a:b\\ \Rightarrow a=b\)

Mà \(a+b=n_{H_2SO_4}=1.0,3=0,3(mol)\)

\(\Rightarrow a=b=0,15(mol)\\ \Rightarrow m_{MgO}=0,15.40=6(g)\)