1.
a)\(\sqrt{300}+\sqrt{243}-\sqrt{192}=10\sqrt{3}+9\sqrt{3}-8\sqrt{3}=11\sqrt{3}\)
b)\(\sqrt{75}-\sqrt{48}+\sqrt{147}=5\sqrt{3}-4\sqrt{3}+7\sqrt{3}=8\sqrt{3}\)
c)\(\dfrac{2}{\sqrt{2}+1}-\dfrac{3}{\sqrt{2}-3}+\dfrac{1}{\sqrt{3}+2}\)
\(=\dfrac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}-\dfrac{3\left(\sqrt{2}+3\right)}{\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)}+\dfrac{1\left(2-\sqrt{3}\right)}{\left(\sqrt{3}+2\right)\left(2-\sqrt{3}\right)}\)
\(=2\sqrt{2}-2+\dfrac{3\sqrt{2}+9}{7}+2-\sqrt{3}\)
\(=\dfrac{14\sqrt{2}-7\sqrt{3}+3\sqrt{2}+9}{7}=\dfrac{17\sqrt{2}-7\sqrt{3}+9}{7}\)
d)\(\dfrac{\left(2+\sqrt{3}\right)\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\sqrt{\dfrac{\left(2+\sqrt{3}\right)^2\left(2-\sqrt{3}\right)}{2+\sqrt{3}}}=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{1}=1\)
2.
a)\(\sqrt{x+2}=4-x\left(x\ge-2\right)\)
\(\Leftrightarrow x+2=\left(4-x\right)^2\)
\(\Leftrightarrow x+2=16-8x+x^2\\ \Leftrightarrow x^2-9x+14=0\\ \Leftrightarrow\left(x-7\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=2\left(N\right)\end{matrix}\right.\)
b)\(\sqrt{x^2+1}=5-x^2\left(x\in R\right)\)
\(\Leftrightarrow x^2+1=25-10x^2+x^4\\ \Leftrightarrow x^4-11x^2+24=0\\ \Leftrightarrow\left(x^2-8\right)\left(x^2-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=8\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
c)\(x^2+\sqrt{x^2-3x+5}=3x+7\left(x\in R\right)\)
\(\Leftrightarrow\left(x^2-3x+5\right)+\sqrt{x^2-3x+5}-12=0\)
Đặt \(\sqrt{x^2-3x+5}=t\left(t\ge\sqrt{\dfrac{11}{4}}\right)\)
\(\Leftrightarrow t+\sqrt{t}-12=0\\ \Leftrightarrow\left[{}\begin{matrix}t=-4\left(L\right)\\t=3\left(N\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2+3x+5}=3\\ \Leftrightarrow x^2+3x+2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
d)\(\sqrt{x+2}-\sqrt{x-6}=2\left(x\ge6\right)\)
\(\Leftrightarrow x+2+x-6-2\sqrt{\left(x+2\right)\left(x-6\right)}=4\\ \Leftrightarrow2x-8-2\sqrt{\left(x+2\right)\left(x-6\right)}=0\\ \Leftrightarrow2\sqrt{\left(x+2\right)\left(x-6\right)}=2x-8\\ \Leftrightarrow\sqrt{\left(x+2\right)\left(x-6\right)}=x-4\\ \Leftrightarrow\left(x+2\right)\left(x-6\right)=x^2-8x+16\\ \Leftrightarrow x^2-4x-12=x^2-8x+16\\ \Leftrightarrow4x=28\Leftrightarrow x=7\left(N\right)\)
Tick nha
giúp em với ạ, cần gấp ạ TvT