\(A=\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3}-\sqrt{5}\)
B=\(B=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8}-2\sqrt{15}\)
\(Cho\sqrt{8-a}+\sqrt{5+a}=5tinh\sqrt{\left(8-a\right)\left(5+a\right)}\)
rút gọn
a) \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
b) \(\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
c) \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
Rút gọn các biểu thức sau:
a) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\) b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
c) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right):\sqrt{6}\) d) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
a)\(\left(\sqrt{3}-\sqrt{2}+1\right).\left(\sqrt{3}-1\right).\)
b)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
c)\(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
d)\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right).\sqrt{5}-\left(3.\sqrt{\frac{1}{10}}+10\right)\)
giúp mk zới:((
Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)
a.\(\sqrt{17}-6\sqrt{2}+3+\sqrt{2 }\)
b.\(\left(3+\sqrt{ }5\right).\left(\sqrt{ }10.\sqrt{ }2\right).\sqrt{3-\sqrt{ }5}\)
c.\(\left(\sqrt{2}-3\right).\sqrt{11+6\sqrt{2}}\)
d.\(\sqrt{23+8\sqrt{7}}-\sqrt{2}\)
nhanh nha gấp lắm trcs 9h
\(a,=\sqrt{17}-5\sqrt{2}+3\\ b,=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\\ =\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=8\\ c,=\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)=2-9=-7\\ d,4+\sqrt{7}-\sqrt{2}\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
Dạng 3.Chứng minh đẳng thức
Bài 1: CM
a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
b)\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Bài 2 :CM
\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{2}}=\sqrt{\sqrt{5}+1}\)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
1. Tính ( rút gọn)
a)\(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
b)\(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
c)\(\sqrt{8+2\sqrt{15}}+\sqrt{\left(\sqrt{2-\sqrt{5}}\right)^2}\)
d)\(\sqrt{12+6\sqrt{3}}.\left(3+\sqrt{3}\right)\)
e) \(\left(2-\sqrt{5}\right).\sqrt{9+4\sqrt{5}}\)
a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
Thu gọn
a) A=\(\dfrac{2\left(\sqrt{2}+\sqrt{6}\right)}{3\sqrt{2+\sqrt{3}}}\) b)B=\(\sqrt{8-2\sqrt{15}}-\sqrt{\left(3-3\sqrt{5}\right)^2}\)
c) C=\(2\sqrt{8\sqrt{3}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}}\)
a: \(A=\dfrac{2\sqrt{2}\left(\sqrt{3}+1\right)}{3\cdot\sqrt{2+\sqrt{3}}}=\dfrac{4\left(\sqrt{3}+1\right)}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{4\left(\sqrt{3}+1\right)}{3\left(\sqrt{3}+1\right)}=\dfrac{4}{3}\)
b: \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\left|3\sqrt{5}-3\right|\)
\(=\sqrt{5}-\sqrt{3}-3\sqrt{5}+3=3-\sqrt{3}-2\sqrt{5}\)