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a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
Lời giải:
a.
\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)
b.
\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)
\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
c.
\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)
\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)
d.
\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)
\(=|2+\sqrt{3}|-|5-\sqrt{3}|=2+\sqrt{3}-(5-\sqrt{3})=-3+2\sqrt{3}\)
a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)
\(=6-2\sqrt{3}+4+3\sqrt{3}\)
\(=10+\sqrt{3}\)
c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=7-5=2
d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=-3+2\sqrt{3}\)
a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=7-5=2\)
d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|-\left(5-\sqrt{3}\right)^2\)
\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)
\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=2\sqrt{3}-3\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
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) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
a, \(=>3-\sqrt{2}+\sqrt{50}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}\)
b, \(=>\dfrac{\sqrt[3]{125.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-4\right).2}=\sqrt[3]{125}-\sqrt[3]{\left(-2\right)^3}\)
\(=5-\left(-2\right)=7\)
c, \(=>\sqrt{6}.\sqrt{\dfrac{6}{2}}-\sqrt{2}-3\sqrt{4.2}=\sqrt{6}.\sqrt{3}-\sqrt{2}-6\sqrt{2}\)
\(=\sqrt{18}-7\sqrt{2}=3\sqrt{2}-7\sqrt{2}=-4\sqrt{2}\)
d, \(=>\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{3-\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{1-\sqrt{3}}{\sqrt{3}-1}=-1\)
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
Khiếp CTV kìa sợ quá ;-;