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1, bình phương x rồi rút gọn ta được
\(x^2=3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{2\left(\sqrt{5}-1\right)\left(\sqrt{5}-2\right)}\)
=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{14-6\sqrt{5}}\)
=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{\left(3-\sqrt{5}\right)^2}\)
=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}\left(3-\sqrt{5}\right)\)
=\(5\sqrt{10}-10\sqrt{2}>0\)
=>x=\(\sqrt{5\sqrt{10}-10\sqrt{2}}\)
a) 7 và \(\sqrt{37}+1\)
=7 và 7,08
=>......
b) \(\sqrt{17}-\sqrt{50}-1\)và \(\sqrt{99}\)
=-3,95 và 9,95
=>.....
a)
= \(\sqrt{18-6\sqrt{6}+3}\)
= \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
= \(\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
= \(|3\sqrt{2}-\sqrt{3}|\)
= \(3\sqrt{2}-\sqrt{3}\)
b)
= \(\sqrt{\frac{7}{2}-\sqrt{7}+\frac{1}{2}}\)
= \(\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)
= \(\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)
= \(|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}|\)
= \(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\)
c)
= \(\sqrt{3+2\sqrt{3}+1}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
d)
Đặt t = \(\sqrt{x-1}\left(ĐK:t\ge0\right)\)
= \(\sqrt{t^2+1-2t}\)
= \(\sqrt{\left(t+1\right)^2}\)
\(=t+1\)
= \(\sqrt{x-1}+1\)
\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2\sqrt{9}\sqrt{6}+3}=\sqrt{\left(\sqrt{18}\right)^2-2\sqrt{18}\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}=\sqrt{3}+3\sqrt{2}\)
\(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{2}}{2}\)
\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
Với \(x\ge1\)thì \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}\sqrt{1}+\left(\sqrt{1}\right)^2}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
T đã tốn mấy phút cuộc đời viết lời giải cho bạn r, tiếc j mấy giây mà bấm k cho t ik =))
a) Ta có :\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\) =\(\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Tương tự : \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\) = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
=>\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)+\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)= \(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)=\(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{3-2}\)=10
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
A =\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(A^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
\(A^2=\left(\sqrt{4-\sqrt{7}}\right)^2-2.\sqrt{4-\sqrt{7}}.\sqrt{4+\sqrt{7}}+\left(\sqrt{4+\sqrt{7}}\right)^2\)
\(A^2=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(A^2=8-2\sqrt{16-7
}\)
\(A^2=8-2\sqrt{9}=8-6=2\)
\(A=\frac{+}{ }\sqrt{2}\)
Vì là biểu thức lên phải có tên . lên mới có A @@!
\(M>0\Leftrightarrow M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}..\)
\(M^2=8-2.\sqrt{16-7}=8-6=3\)
\(M=\sqrt{3}.\)
bẹn Nguyễn Thị Thùy Dương ơi, 8 - 6 =3 là sai r đó nha