1) Rút gọn biểu thức
a) \(\sqrt{6+2\sqrt{5}}\)+ \(\sqrt{6-2\sqrt{5}}\)
b) \(\sqrt{3+2\sqrt{2}}\)+ \(\sqrt{6-4\sqrt{2}}\)
c) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
d) \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
2) Phân tích thành nhân tử
a) x-1 ( x\(\ge\)0)
b) x + 5\(\sqrt{2}\) + 6 ( x \(\ge\)0 )
c) x-4 ( x\(\ge\)0 )
1.
\(a.\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}=\sqrt{2}+1+2-\sqrt{2}=3\)\(c.\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
\(d.\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
2.
\(a.x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b.x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)( mạo danh sửa đề)
\(c.x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(1a.\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}=\sqrt{2}+1+2-\sqrt{2}=3\)\(c.\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2.3\sqrt{2}+2}-\sqrt{9-2.3\sqrt{2}+2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)\(d.\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}=\sqrt{2}\)\(2a.x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(b.x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
\(c.x-4=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
