a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)

Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)

\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

Nhận xét: \(\text{ *)}\) Nếu  \(x+y+z=0\)  thì  \(x^3+y^3+z^3=3xyz\)     

Thật vậy,  từ  \(x+y+z=0\)

Suy ra:  \(x+y=-z\)  \(\left(\text{*}\right)\)

\(\Leftrightarrow\)  \(\left(x+y\right)^3=\left(-z\right)^3\)  

\(\Leftrightarrow\)  \(x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)

\(\Leftrightarrow\)  \(x^3+y^3+z^3=-3x^2y-3xy^2\)

\(\Leftrightarrow\)  \(x^3+y^3+z^3=-3xy\left(x+y\right)\)

\(\Leftrightarrow\)  \(x^3+y^3+z^3=3xyz\)  (theo \(\left(\text{*}\right)\)  )

                                                              \(-------------\)

Theo giả thiết, ta có:

\(a+b+c=0\)

\(\Leftrightarrow\)  \(b+c=-a\)

\(\Leftrightarrow\)  \(\left(b+c\right)^2=\left(-a\right)^2\)

\(\Leftrightarrow\)  \(b^2+2bc+c^2=a^2\)

\(\Leftrightarrow\)  \(2bc=a^2-b^2-c^2\)

Tương tự, ta cũng có  \(2ac=b^2-a^2-c^2\)  \(;\) \(2ab=c^2-a^2-b^2\)

Mặt khác,  vì \(a+b+c=0\)  nên  \(a^3+b^3+c^3=3abc\)  (theo nhận xét trên)

Do đó,  \(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3}{2abc}+\frac{b^3}{2abc}+\frac{c^3}{2abc}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)  (do  \(abc\ne0\)  

tu a + b + c = 0 suy ra a= - (b+c) suy ra a^2 = (b+c)^2=b^2 +c^2 + 2bc                                                                                                    suy ra a^2 - b^2 - c^2 =2bc . tuong tu ta cung co b^2-a^2-c^2=2ac ; c^2- a^2 -b^2=2ab                                                                          do do A = a^2/2bc + b^2/2ac+c^2/2ab =a^3/2abc+b^3/2abc +c^3/2abc                                                                                                           lai co a+b+c=o nen a+b=-c suyra a^3+b^3+3ab(a+b)= -c^3 do do a^3 +b^3 +c^3=3abc                                                                     vay A=3abc/2abc=3/2 (abc khac 0 : a+b=c=o)

bài này có trong câu hỏi tương tự nhé bạn

Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)

\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)

Thay (*) vào M ta được:

\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)

\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)

\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)

\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)

Vậy M = 0

\(=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2a+2b+2c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(A=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

Linh không biết a + b + c = 0 để làm gì?

Có a + b + c = 0

=> a + b = - c

=> (a + b)² = c²

=> a² + b² + 2ab = c²

=> a² + b² - c² = - 2ab

Tương tự, b² + c² - a² = - 2bc và c² + a² - b² = - 2ca

Do đó \(A=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)

a+b+c=0=>a+b=-c=>a²+b²+2ab=c²=>a²+b²-c²=-2ab

Tương tự b²+c²-a²=-2bc,c²+a²-b²=-2ac

=>\(A=\frac{-ab}{2ab}+\frac{-bc}{2bc}+\frac{-ca}{2ca}=\frac{-3}{2}\)