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mình nghĩ bài này sai đề,
ĐÚng phải là\(\sqrt[3]{2+\sqrt{3}}\)
( KHÔNG CHẮC NỮA :D )
Bài làm:
a) \(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}=\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{2}=\frac{\sqrt{14}-\sqrt{2}}{2}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (đề vậy chứ)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2\)
c) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
d) \(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
a)ĐKXĐ : x > 0
P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)
c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)
Từ (*) <=> a2 -(25a2 -40a + 16 ) =4
<=> -24a2 + 40a - 20 = 0
=> \(\Delta'=-80< 0\)
=> PT vô nghiệm
=> ko tồn tại x thỏa mãn