mng ới ;-;

Bài 1 : 

\(\Delta'=m^2-2\left(m-2018\right)=m^2-2m+2018=\left(m-1\right)^2+2017>0\forall m\)

Vậy pt luôn có 2 nghiệm phân biệt 

a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)

  \(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)

  \(=0\)

b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)

Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)

\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)

Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

a)C=\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{6}}\)   -\(\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{6}}\)

=\(\frac{1+\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{1-\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

=\(\frac{1+\sqrt{6}-1+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

=\(\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

chịu 😅

, \(A=\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

\(=\frac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{3}-\frac{5\left(4-\sqrt{7}\right)}{9}\)

\(=\frac{-16+4\sqrt{7}}{4}+\frac{18\sqrt{7}+36-20+5\sqrt{7}}{9}=-4+\sqrt{7}+\frac{23\sqrt{7}+16}{9}\)

b,\(B=\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{2}+\frac{5\sqrt{6}}{6}\)

\(=\frac{12\sqrt{6}+5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)

a, \(\frac{a}{\sqrt{a}}=\sqrt{a}\)

b, \(\frac{a}{\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)

c, \(\frac{x}{\sqrt{3x^3}}=\frac{x}{x\sqrt{3x}}=\frac{1}{\sqrt{3x}}=\frac{\sqrt{3x}}{3x}\)

d, \(\frac{4y^2}{\sqrt{2y^5}}=\frac{4y^2}{y^2\sqrt{2y}}=\frac{4}{\sqrt{2y}}=\frac{4\sqrt{2y}}{2y}=\frac{2\sqrt{2y}}{y}\)

a)\(\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}\)                b) \(\dfrac{a}{\sqrt{ab}}=\dfrac{a\sqrt{ab}}{\left(\sqrt{ab}\right)^2}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b}\)                                              c) \(\dfrac{x}{\sqrt{3x^3}}=\dfrac{x\sqrt{3x}}{\sqrt{3x^3.\sqrt{3x}}}=\dfrac{x\sqrt{3x}}{\left(\sqrt{3x^2}\right)^2}=\dfrac{x\sqrt{3x}}{\left(3x^2\right)^2}=\dfrac{x\sqrt{3x}}{3x^2}=\dfrac{\sqrt{3x}}{3x}\)    

a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\)  =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\)  -  \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+  \(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}.21a}\) -  \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\)  -   \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+  \(\sqrt{21a}\)

=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\)  +  \(\sqrt{21a}\)

=\(\frac{-10}{21}\sqrt{21a}\)

b)

N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)

=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)

=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)

=\(\frac{1}{6}\sqrt{6x}\)

em lớp 8 nene làm theo cách hiểu thôi ạ

Answer:

\((ac+bd)^2+(ad-bc)^2\)

\(=(ac)^2+(2acbd)+(bd)^2+(ad)^2-(2acbd)+(bc)^2\)

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

\(=a^2(c^2+d^2)+b^2(c^2+d^2)\)

\(=(c^2+d^2)(a^2+b^2)\)

tự làm