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\(A=\sqrt{9.7}-2\sqrt{25.7}+\sqrt{9.7.4}-\frac{1}{7}\sqrt{4.7}\)
\(=3\sqrt{7}-10\sqrt{7}+6\sqrt{7}-\frac{2}{7}\sqrt{7}\)
\(=\frac{-9}{7}\sqrt{7}\)
Nếu đúng tk nhé
a = \(\sqrt{63}-2\sqrt{175}+\sqrt{252}-\frac{1}{7}\sqrt{28}\)
= \(\sqrt{\frac{4}{7}}\left(1,5-5+3-1\right)\)
= \(-1,5\sqrt{\frac{4}{7}}\)
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=\left(6-4+50-63\right)\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=\left(2+9-30-\dfrac{6}{5}\right)\sqrt{7}=-20,2\sqrt{7}\)\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=10\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-12\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=-\dfrac{101}{5}\sqrt{7}\)
\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=20\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-2\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
\(a.\sqrt{200}-\sqrt{32}+\sqrt{72}=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
\(b.\sqrt{175}-\sqrt{112}+\sqrt{63}=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
\(c.4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
\(A=4-7+6=3\)
\(B=\sqrt{4.7^2}-2\sqrt{25.7^2}+\sqrt{9.7^2}=2.7-2.5.7+3.7=-35\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}.4=4\left(\sqrt{3}-1\right)=4\sqrt{3}-4\)
\(D=\sqrt{\left(2+\sqrt{3}\right)^2}.\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(1+\sqrt{2}\)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)
\(A=...\)
\(=3\sqrt{7}-2.5\sqrt{7}+6\sqrt{7}-\dfrac{1}{7}.2\sqrt{7}\)
\(=\left(3-2.5+6-\dfrac{1}{7}.2\right)\sqrt{7}\)
\(=-\dfrac{9\sqrt{7}}{7}\)