a ) \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)

     = \(\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

    = \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

 = \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

= \(\frac{-\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{-1}{\left(x-z\right)\left(y-z\right)}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)

Chúc bạn học tốt !!!

\(a^3+b^3⋮3\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)⋮3\)

\(+,a^2-ab+b^2⋮3\Leftrightarrow a^2+2ab+b^2⋮3\Leftrightarrow\left(a+b\right)^2⋮3\Rightarrow a+b⋮3\)

\(\Rightarrow dpcm\)

Bài 1 

B C H A K

Ta có : \(S_{ABC}=\frac{1}{2}AB.CK=\frac{1}{2}AC.BH\)

Suy ra :  \(AB.CK=AC.BH\Rightarrow\frac{BH}{CK}=\frac{AB}{AC}\)

Mà AB = 3AC ( gt )

\(\Rightarrow\frac{BH}{CK}=\frac{3AC}{AC}=3\)

Vậy đường cao BH dài gấp 3 lần đường cao CK .

Bài 2 

A B C D I

B và H đối xứng qua AD.

I và A đối xứng với chính nó qua AD

Nên \(\widehat{AIB}\) đối xứng với \(\widehat{AIH}\) qua AD 

\(\Rightarrow\widehat{AIB}=\widehat{AIH}\)

\(\widehat{AIB}=\widehat{DIC}\) ( đối đỉnh )

\(\Rightarrow\widehat{AIB}=\widehat{DIC}\)

Vậy \(\widehat{AIB}=\widehat{DIC}\)

Chúc bạn học tốt !!!

\(M=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)

\(=2\left(a+b\right)\left(a^2+ab+b^2\right)-3a^2-3b^2\)

\(=2a^2+2ab+2b^2-3a^2-3b^2\)

\(=-a^2+2ab-b^2\)

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

\(=-\left(a-b\right)^2\)

\(=-\left(1-b-b\right)^2=-\left(1-2b\right)^2\)

A=((x-3)+(x+1))^2>=0

A=(x-2)^2>=0

Dấu bằng xảy ra khi

(x-2)^2=0

x-2=0

x=0+2

x=2

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow2A=3^{32}-1\)

\(\Leftrightarrow A=3^{31}-\frac{1}{2}\)

A=\(413^2-413.26+13^2=413^2-2.413.13+13^2\)

=> A=\(\left(413-13\right)^2=400^2=160000\)

chuc ban hoc tot