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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...
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 \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=2\sqrt{3}-3\)
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}\)
\(=\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\)
\(=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)
\(A=\sqrt{13+4\sqrt{10}}=\sqrt{13+2\sqrt{40}}=\sqrt{8+2.\sqrt{5}.\sqrt{8}+5}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=\sqrt{8}+\sqrt{5}\)
\(B=\sqrt{46-6\sqrt{5}}=\sqrt{46-2\sqrt{45}}=\sqrt{\left(\sqrt{45}-1\right)^2}=\sqrt{45}-1=3\sqrt{5}-1\)
\(C=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{5}+\sqrt{3}}{2}-\dfrac{\sqrt{7}+\sqrt{5}}{2}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{7}}{2}\)
\(C=\dfrac{-2\sqrt{3}-2\sqrt{2}+\sqrt{3}-\sqrt{7}}{2}=\dfrac{-\sqrt{3}-2\sqrt{2}-\sqrt{7}}{2}\)
Nga Văn sr thiếu vế :3