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h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)
b) \(B=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
c) \(C=\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}+1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}+1+2-\sqrt{2}=3\)
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)
\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)
\(=6+\sqrt{5}\)
\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)
\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)
\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}-2+\sqrt{5}\)
\(=-3+\sqrt{3}+\sqrt{5}\)
\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{3}\)
\(=\sqrt{3}-1+2\sqrt{3}\)
\(=-1+3\sqrt{3}\)
\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\sqrt{6}\)
\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)
Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)
Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
.... Đúng thì ủng hộ nha ....
Kết bạn với mình ... ;) ;)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(A=\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)+\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(A=\frac{2\sqrt{3}-2+3-\sqrt{3}+2\sqrt{3}+2-3-\sqrt{3}}{\sqrt{3}\left(3-1\right)}\)
\(A=\frac{2\sqrt{3}}{2\sqrt{3}}=1\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}.\)
\(=\frac{\left(2+\sqrt{3}\right)\left(2-\sqrt{4+2\sqrt{3}}\right)}{\left(2+\sqrt{4+2\sqrt{3}}\right)\left(2-\sqrt{4+2\sqrt{3}}\right)}+\frac{\left(2-\sqrt{3}\right)\left(2+\sqrt{4-2\sqrt{3}}\right)}{\left(2-\sqrt{4-2\sqrt{3}}\right)\left(2+\sqrt{4-2\sqrt{3}}\right)}\)
\(=\frac{4-2\sqrt{4+2\sqrt{3}}+2\sqrt{3}-\sqrt{3\left(4+2\sqrt{3}\right)}}{4-4-2\sqrt{3}}+\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}}{4-4+2\sqrt{3}}\)
\(=\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}}{2\sqrt{3}}-\frac{4-2\sqrt{4+2\sqrt{3}}+2\sqrt{3}-\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)
\(=\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}-4+2\sqrt{4+2\sqrt{3}}-2\sqrt{3}+\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)
\(=\frac{2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}+2\sqrt{4+2\sqrt{3}}-2\sqrt{3}+\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)