\(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(\sqrt{8-2\sqrt{7}}+\sqrt{23-8\sqrt{7}}\)
\(\sqrt{8-2\sqrt{7}}+\sqrt{23-8\sqrt{7}}\)
=\(\sqrt{\sqrt{7}^2-2.\sqrt{7}.1+1^2}+\sqrt{4^2-2.4.\sqrt{7}+\sqrt{7}^2}\)
=\(\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(4-\sqrt{7}\right)^2}\)
=\(\left|\sqrt{7}-1\right|+\left|4-\sqrt{7}\right|\)
=\(\sqrt{7}-1+4-\sqrt{7}\)
=3
\(\sqrt{8-2\sqrt{7}}\) + \(\sqrt{23-8\sqrt{7}}\)
= \(\sqrt{7-2\sqrt{7}+1}\) + \(\sqrt{16-8\sqrt{7}+7}\)
= \(\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7}+1^2}\) + \(\sqrt{4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2}\)
= \(\sqrt{\left(\sqrt{7}-1\right)^2}\) + \(\sqrt{\left(4-\sqrt{7}\right)^2}\)
= \(\left|\sqrt{7}-1\right|\) + \(\left|4-\sqrt{7}\right|\)
= \(-\)( \(\sqrt{7}-1\)) + ( \(4-\sqrt{7}\)) ( vì \(\sqrt{7}-1\)< 0; \(4-\sqrt{7}\)>0)
= \(-1+\sqrt{7}\) + \(4-\sqrt{7}\)
= 3
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
đề là rút gọn các biểu thức sau
nhờ mọi người giải giúp mình. cảm ơn mn nhìu
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)
=2căn 5-2-2căn 5
=-2
d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)
Bài 1: rút gọn biểu thức
d)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
a)\(\sqrt{9-4\sqrt{5}}+\sqrt{5}\)
b)\(\sqrt{6-4\sqrt{2}}+\sqrt{2}\)
c)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
1)d) \(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{4^2+2.4.\sqrt{7}+\sqrt{7^2}}-\sqrt{7}\)
\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(=4+\sqrt{7}-\sqrt{7}\)
\(=4\)
a) \(\sqrt{9-4\sqrt{5}}+\sqrt{5}\)
=\(\sqrt{\left(\sqrt{2}\right)^2-2.2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{5}\)
=\(\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{5}\)
=\(\left|\sqrt{2}-\sqrt{5}\right|+\sqrt{5}\)
=\(\sqrt{2}-\sqrt{5}+\sqrt{5}\)
=\(\sqrt{2}\)
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
rút gọn biểu thức:
\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}\)
\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}=\) \(\sqrt{1-2\sqrt{7}+7}-\sqrt{7-2.4.\sqrt{7}+16}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-4\right)^2}\)
\(=\sqrt{7}-1-\left(-\sqrt{7}+4\right)\)
\(=\sqrt{7}-1+\sqrt{7}-4\)\(=2\sqrt{7}-5\)
chúc bn học tốt
=\(\sqrt{\left(\sqrt{7}-1\right)^2}\)- \(\sqrt{\left(4-\sqrt{7}\right)^2}\)
= \(\sqrt{7}\)- 1 - 4 + \(\sqrt{7}\)
= \(2\sqrt{7}\)-5
đ/á ra hơi kì
#mã mã#
mình nghĩ là hai bn làm đúng đó
Rút gọn: \(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
giải gấp giúp e nha
Tìm x
d, \(\sqrt{x-2\sqrt{x-1}=\sqrt{x-1}-1}\)
e, \(\sqrt{1-12x+36x^2}=5\)
g, \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
- Đề sai nhiều vậy sửa lại đi bạn ;-;
e) Ta có: \(\sqrt{1-12x+36x^2}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{2}{3}\right\}\)
Chứng minh :
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
d) \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
a) \(9+4\sqrt{5}=4+4\sqrt{5}+5=2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{5}+2\right)^2\left(ĐPCM\right)\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\left(đpcm\right)\)
b)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(đpcm\right)\)
c)\(\left(4-\sqrt{7}\right)^2=16-8\sqrt{7}+7=23-8\sqrt{7}\left(đpcm\right)\)
d)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}=4+\sqrt{7}-\sqrt{7}=4\left(đpcm\right)\)
\(\sqrt{23-8\sqrt{ }7}\)
\(\sqrt{23-8\sqrt{7}}=\sqrt{\left(\sqrt{16}-\sqrt{7}\right)^2}=\sqrt{16}-\sqrt{7}=4-\sqrt{7}\)
\(\sqrt{23-8\sqrt{7}}=4-\sqrt{7}\)