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\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
+) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
\(=3\sqrt{4.5}-2\sqrt{9.5}+4\sqrt{5}\)
\(=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
\(=4\sqrt{5}\)
+) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=\left(2\sqrt{7}-\sqrt{28}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=\left(2\sqrt{7}-2\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=7+7\sqrt{8}\)
\(A=\sqrt{24+8\sqrt{5}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{5+2.4\sqrt{5}+16}+\sqrt{4-2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}+4\right)}^2+\sqrt{\left(2-\sqrt{3}\right)}^2\)
\(=|\sqrt{5}+4|+|2-\sqrt{3}|\)
\(=\sqrt{5}+4+4-\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}+8\)
Ko biết đề sai ko?
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
con cacacacacacacacacacacacacacacacacacca
@@22@22@22@@222@@2@@2@@@2@2
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2}{2}\)
c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))
\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-3}{3}\)
b) Tử \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\) (hằng đăngt thức số 3 )
a) đặt A = \(\sqrt{14+8\sqrt{3}}.\left(2\sqrt{2}+\sqrt{3}\right)\)
=> \(A^2=\left(14+8\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(=\left(14+8\sqrt{3}\right)\left(14+8\sqrt{3}\right)\)
\(=\left(14+8\sqrt{3}\right)^2\)
=> A = \(14+8\sqrt{3}\)
b) đặt B = \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
=> \(B^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
= \(4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
= \(8-2\sqrt{9}\)
\(=8-6=2\)
=> C = \(\sqrt{2}\)
c) đặt C = \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
=> \(C^2=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+3+\sqrt{5}\)
\(=6+2\sqrt{1}\) \(=8\)
=> C = \(\sqrt{8}\)
mong bài mk đúng :)~~