\(\frac{-3+\sqrt{17}}{4}\frac{-3-\sqrt{17}}{4}\)
Những câu hỏi liên quan
Tính: \(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
Ta có:
\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)
\(\Leftrightarrow A^2=10-2\sqrt{25-17}=10-4\sqrt{2}\)
\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)
Ta lại có:
\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)
\(\Leftrightarrow B=\sqrt{2}\)
Thế vô biểu thức ban đầu ta được
\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)
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ta có :
\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)
\(\Leftrightarrow A^2=10-2\sqrt{25-17=10-4\sqrt{2}}\)
\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)
ta lại có :
\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)
\(\Leftrightarrow B=\sqrt{2}\)
the vo bieu thuc ban dau ta duoc
\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2=\sqrt{2}}\)
\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)
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Xem thêm câu trả lời
\(\frac{6-2\sqrt{3}}{\sqrt{3}-1}-\frac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{17+4\sqrt{15}}\)
Thực hiện phép tính :
A = \(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
Tinh
a, \(3\sqrt{50}-2\sqrt{98}-5\sqrt{18}-\sqrt{63}-2\sqrt{28}\)
b, \(\sqrt{42-10\sqrt{17}}+\sqrt{3-8\sqrt{17}}\)
c, \(\frac{4}{\sqrt{3}+1}+\frac{6}{\sqrt{3}-3}-\frac{5}{\sqrt{3}-2}\)
1/ frac{2}{3-sqrt{7}}sqrt{frac{6sqrt{2}-2sqrt{14}}{3sqrt{2}+sqrt{14}}}2/ sqrt{6+2sqrt{sqrt{5}-sqrt{13-sqrt{48}}}}3/ frac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}4/ frac{24}{sqrt{7}+1}+frac{4}{3+sqrt{7}}-frac{3}{sqrt{7}+2}left(4-sqrt{7}right)5/ sqrt{7-3sqrt{5}}left(7+3sqrt{5}right)left(3sqrt{2}+sqrt{10}right)
Đọc tiếp
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
x^2-left(2m+3right)x+m^2+3m+20.left{x^2-left(2m+3right)x+frac{left(2m+3right)^2}{4}right}frac{left(2m+3right)^2+4m^2+12m+8}{4}left(x-frac{2m+3}{2}right)^2frac{8m^2+24m+17}{4}Leftrightarrowhept{begin{cases}2x-2m+3sqrt{8m^2+24m+17}2x-2m+3-sqrt{8m^2+24m+17}end{cases}}để căn có nghĩa thì8m^2+24m+17left(m^2+3m+frac{9}{4}right)-frac{1}{8}ge0left(m+frac{3}{2}right)^2gefrac{1}{8} suy ra m.....vậy pt có 2 nghiệm phân biệt với m.....Leftrightarrowhept{begin{cases}x1frac{1}{2}sqrt{8m^2+24+17}+m-frac{3}{2}...
Đọc tiếp
\(x^2-\left(2m+3\right)x+m^2+3m+2=0.\)
\(\left\{x^2-\left(2m+3\right)x+\frac{\left(2m+3\right)^2}{4}\right\}=\frac{\left(2m+3\right)^2+4m^2+12m+8}{4}\)
\(\left(x-\frac{2m+3}{2}\right)^2=\frac{8m^2+24m+17}{4}\)
\(\Leftrightarrow\hept{\begin{cases}2x-2m+3=\sqrt{8m^2+24m+17}\\2x-2m+3=-\sqrt{8m^2+24m+17}\end{cases}}\)
để căn có nghĩa thì
\(8m^2+24m+17=\left(m^2+3m+\frac{9}{4}\right)-\frac{1}{8}\ge0\)
\(\left(m+\frac{3}{2}\right)^2\ge\frac{1}{8}\) " suy ra m.....
vậy pt có 2 nghiệm phân biệt với m.....
\(\Leftrightarrow\hept{\begin{cases}x1=\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\\x2=-\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\end{cases}}\)
\(x1< -3\Leftrightarrow-3< \frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\)
\(\Leftrightarrow m>-3-\frac{1}{2}\sqrt{8m^2+24+17}+\frac{3}{2}\)
\(x1< x2\Leftrightarrow\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}< -\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\)
\(\Leftrightarrow0< -\sqrt{8m^2+24+17}\)
\(x2< 6\Leftrightarrow-\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}< 6\)
\(\Leftrightarrow m< 6+\frac{1}{2}\sqrt{8m^2+24+17}+\frac{3}{2}\)
dcpcm =))
tính A=\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^3\)
B=\(\sqrt{\frac{3-2\sqrt{2}}{17-12\sqrt{2}}}-\sqrt{\frac{3+2\sqrt{2}}{17+12\sqrt{2}}}\)
Chứng minh rằng \(17< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}< 18\)
+ \(2\cdot\frac{1}{\sqrt{n}+\sqrt{n+1}}< \frac{2}{\sqrt{n}+\sqrt{n}}< 2\cdot\frac{1}{\sqrt{n-1}+\sqrt{n}}\) \(\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(\Rightarrow A>2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\right)\)
\(\Rightarrow A>2\left(\sqrt{101}-\sqrt{2}\right)>17\)
+ \(A< 2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\Rightarrow A< 2\left(\sqrt{100}-1\right)=18\)
Chứng minh rằng \(17< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}< 18\)