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a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\dfrac{\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{5}}{\sqrt{7}\sqrt{3}+\sqrt{7}\sqrt{5}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=3\dfrac{3\sqrt{3}+3\sqrt{5}}{3\sqrt{3}+3\sqrt{5}}=3.1=3\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)-\sqrt{3}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1-\sqrt{3}\)
P/s: bạn làm thêm bước nữa nha, mình lười, hehe
d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}}{\sqrt{5}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{5}-1}=\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
\(a,\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
\(=2+2\sqrt{2}-2\sqrt{2}\)
\(=2\)
\(b,\sqrt{\left(\sqrt{5}-\sqrt{10}\right)^2}-\sqrt{10.\left(\sqrt{2+1}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{10}\right|-\sqrt{10.3}\)
\(=\left(\sqrt{10}-\sqrt{5}\right)-\sqrt{30}\)
\(=\sqrt{10}-\sqrt{5}-\sqrt{30}\)
\(=\sqrt{5}.\left(\sqrt{2}-1-\sqrt{6}\right)\)
\(c,\left(\sqrt{2}+\sqrt{3}\right)^2.\sqrt{49-20\sqrt{6}}\)
\(=\left(2+3+2.\sqrt{2}.\sqrt{3}\right).\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)\)
\(=1\)
d,\(\dfrac{2}{\sqrt{8-\sqrt{60}}}-\sqrt{\dfrac{\sqrt{18}+\sqrt{27}}{\sqrt{3}+\sqrt{2}}}\)
\(=\dfrac{2}{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}-\sqrt{\dfrac{3\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{2}}}\)
\(=\dfrac{2}{\sqrt{5}-\sqrt{3}}-\sqrt{\dfrac{3.\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{3}+\sqrt{2}}}\)
\(=\dfrac{2}{\sqrt{5}-\sqrt{3}}-\sqrt{3}\)
\(=\dfrac{2.\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\sqrt{3}\)
\(=\dfrac{2\sqrt{5}+2\sqrt{3}}{2}-\sqrt{3}\)
\(=\dfrac{2\sqrt{5}+2\sqrt{3}-2\sqrt{3}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}\)
\(=\sqrt{5}\)
Lời giải:
Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)
Khi đó: \(D=a+b\)
\(\left\{\begin{matrix} a^3+b^3=2+10\sqrt{\frac{1}{27}}+2-10\sqrt{\frac{1}{27}}=4\\ ab=\sqrt[3]{(2+10\sqrt{\frac{1}{27}})(2-10\sqrt{\frac{1}{27}})}=\frac{2}{3}\end{matrix}\right.\)
Có:
\(a^3+b^3=(a+b)^3-3ab(a+b)\)
\(\Leftrightarrow 4=D^3-3.\frac{2}{3}D\)
\(\Leftrightarrow D^3-2D-4=0\)
\(\Leftrightarrow (D-2)(D^2+2D+2)=0\)
Vì \(D^2+2D+2=(D+1)^2+1\neq 0\), do đó \(D-2=0\Leftrightarrow D=2\)