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\(\sqrt{12-6\sqrt{3}}-\)\(\sqrt{7-4\sqrt{3}}=\)\(\sqrt{3^2-2.3.\sqrt{3}+3}-\)\(\sqrt{2^2-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=3-\sqrt{3}-\)\(\left(2-\sqrt{3}\right)\)
\(=3-\sqrt{3}-2+\sqrt{3}\)\(=1\)
\(\frac{\sqrt{3}+\sqrt{7}}{\sqrt{3}-\sqrt{7}}+\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)+\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)^2+\left(\sqrt{3}-\sqrt{7}\right)^2}{3-7}\)
\(=\frac{3+2\sqrt{3}.\sqrt{7}+7+3-2\sqrt{3}.\sqrt{7}+7}{-4}\)
\(=\frac{3+7+3+7}{-4}\)
\(=\frac{20}{-4}=-5\)
\(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)
=>\(A^2=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{4}\)
=>A^2=2căn 7-4
=>A=2căn 7-4
=>\(M=\dfrac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Ta có:
\(x^3=6+3x.\sqrt[3]{9-8}\Leftrightarrow x^3-3x=6\)
\(y^3=34+3y\sqrt[3]{17^2-12^2.2}\Leftrightarrow y^3-3y=34\)
=>B = 6 + 34 + 2017 =2057
Ta có:
x3=6+3x.3√9−8⇔x3−3x=6
y3=34+3y3√172−122.2⇔y3−3y=34
Nên ta suy ra được => B = 6 + 34 + 2017 =2057
Chúc bạn học tốt :)))
\(\sqrt{12-3}.\sqrt{7}-\sqrt{12+3}.\sqrt{7}\)
\(=\sqrt{7}.\sqrt{12^2-3^2}\)
\(=\sqrt{7}.\sqrt{135}\)
\(=\sqrt{945}\)
\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
Ta có :
\(\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)^2\)
= \(12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{12-3\sqrt{7}}.\sqrt{12+3\sqrt{7}}\)
= \(24-2.\sqrt{12^2-\left(3\sqrt{7}\right)^2}\)
= \(24-2.\sqrt{144-63}\)
= \(24-18=6\)
Mặt khác ta dễ thấy : \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}< 0\)
\(\Rightarrow\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=-\sqrt{6}\)
Chúc bạn học tốt !!!