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11 tháng 6 2017

a, \(5\sqrt{\left(-2\right)^4}=5\sqrt{2^4}=5.2^2=5.4=20\)

b, \(-4\sqrt{\left(-3\right)^6}=-4\sqrt{3^6}=-4.3^3=-4.27=-108\)

c,\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{5^8}}=\sqrt{5^4}=5^2=25\)

d ,\(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\)

\(=2\sqrt{5^6}+3\sqrt{2^8}\)

=\(2.5^3+3.2^4=2.125+3.16=298\)

23 tháng 6 2018

a) \(5\sqrt{\left(-2\right)^4}\) \(=5\left|\left(-2\right)^2\right|=5.4=20\)

b) \(-4\sqrt{\left(-3\right)^6}=-4\left|\left(-3\right)^3\right|=-4.27=-108\)

c) \(\sqrt{\sqrt{\left(-5\right)^8}}=\left|\left(-5\right)^4\right|=5^4=625\)

d) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\) \(=2\left|\left(-5\right)^3\right|+3\left|\left(-2\right)^4\right|\)

\(=-2.\left(-125\right)+3.16\)

\(= 250 + 48 = 298\)

NV
7 tháng 5 2023

\(5\sqrt{\left(-2\right)^4}=5\sqrt{4^2}=5.4=20\)

\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{27^2}=-4.27=-108\)

\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{5^4}=\sqrt{25^2}=25\)

7 tháng 5 2023

cảm ơn thầy ạ

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

AH
Akai Haruma
Giáo viên
5 tháng 10 2021

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}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

4 tháng 7 2021

a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)

\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)

\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)

c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)

\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)

d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)

 

22 tháng 6 2018

a) \(5\sqrt{\left(-2\right)^4}=5\sqrt{\left(2^2\right)^2}=5\left|4\right|=5.4=20\)

22 tháng 6 2018

b)\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{\left(3^3\right)^2}=-4\left|27\right|=-4.27=-108\)

c) \(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{\left(5^2\right)^2}=25\)

d)

30 tháng 5 2023

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

1 tháng 10 2023

\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)

\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)

\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)

\(A=2^2-\left(\sqrt{5}\right)^2\)

\(A=4-5\)

\(A=-1\)

____

\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(B=6-121\)

\(B=-115\)

27 tháng 8 2017

a) \(VT=2\sqrt{6}-4\sqrt{2}+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

\(=-4\sqrt{2}+1+4\sqrt{2}+8\)

\(=1+8\)

\(=9\)

\(\Rightarrow VT=VP\) (đpcm).

b) \(VT=\left(3\sqrt{10}-3\sqrt{2}+\sqrt{50}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)

\(=\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)

\(=\left(2\sqrt{10}-2\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(40+8\sqrt{20}+8\right)\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(48+16\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\sqrt{16\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\sqrt{16\left(9-5\right)}\)

\(=\sqrt{64}\)

\(=8\)

\(\Rightarrow VT=VP\) (đpcm).

c) \(VT=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}\)

\(=2\left(\sqrt{5}+2\right)-\dfrac{2\left(2-\sqrt{5}\right)}{-1}\)

\(=2\sqrt{5}+4+2\left(2-\sqrt{5}\right)\)

\(=2\sqrt{5}+4+4-2\sqrt{5}\)

\(=4+4\)

\(=8\)

\(\Rightarrow VT=VP\) (đpcm).