\(\sqrt{12-3}.\sqrt{7}-\sqrt{12+3}.\sqrt{7}\)

\(=\sqrt{7}.\sqrt{12^2-3^2}\)

\(=\sqrt{7}.\sqrt{135}\)

\(=\sqrt{945}\)

\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

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 !!!

\(\sqrt{12-6\sqrt{3}}-\)\(\sqrt{7-4\sqrt{3}}=\)\(\sqrt{3^2-2.3.\sqrt{3}+3}-\)\(\sqrt{2^2-2.2.\sqrt{3}+3}\)

\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=3-\sqrt{3}-\)\(\left(2-\sqrt{3}\right)\)

\(=3-\sqrt{3}-2+\sqrt{3}\)\(=1\)

các bạn ơi giúp mình vs 

Rút gon P=\((-3x +5\sqrt(x)-2)/(x+4\sqrt(x)-5) \)

A. -0,8 ×0,125=-0,1

b. 2^3+3^2=8+9=17

c.=1

d.=-2

\(\frac{\sqrt{3}+\sqrt{7}}{\sqrt{3}-\sqrt{7}}+\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}+\sqrt{7}}\)

\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)+\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)^2+\left(\sqrt{3}-\sqrt{7}\right)^2}{3-7}\)

\(=\frac{3+2\sqrt{3}.\sqrt{7}+7+3-2\sqrt{3}.\sqrt{7}+7}{-4}\)

\(=\frac{3+7+3+7}{-4}\)

\(=\frac{20}{-4}=-5\)

Bài này đơn giản chỉ quy đồng về HDT thoi

\(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)

=>\(A^2=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{4}\)

=>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 :)))