Giải : 

a³ + b³ + a²c + b²c - abc

= ( a³ + b³ ) + ( a²c + b²c - abc )

= ( a + b ) ( a² - ab + b² ) + c ( a² - ab + b² ) 

= ( a² - ab + b² ) ( a + b + c )

Vì a + b + c = 0 , nên ( a + b + c  ) ( a² - ab + b² ) = 0

Do đó a³ + b³+ a²c + b²c - abc = 0

=a ^3+a^2c+a^2b-a^2b-abc+b^2c+b^3+b^2a-b^2a =a^2(a+b+c)-a^2b-abc+b^2(a+b+c)-b^2a = -a^2b-abc-b^2a = -ab(a+b+c)=-ab 0 =0 vậy đa thức này bằng 0 

bang 0

Ta có

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

\(=a\left(a+1\right)+b\left(b+1\right)+c\left(c+1\right)+d\left(d+1\right)\)

Ta thấy

\(a\left(a+1\right);b\left(b+1\right);c\left(c+1\right);d\left(d+1\right)\) là tích của 2 số tự nhiên liên tiếp nên các tích trên đều chia hết cho 2

\(\Rightarrow a\left(a+1\right)+b\left(b+1\right)+c\left(c+1\right)+d\left(d+1\right)⋮2\)

\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)+\left(a+b+c+d\right)⋮2\)

Ta có

\(a^2+c^2=b^2+d^2\Rightarrow\left(a^2+b^2+c^2+d^2\right)=2\left(b^2+d^2\right)⋮2\)

\(\Rightarrow a^2+b^2+c^2+d^2⋮2\)

\(\Rightarrow a+b+c+d⋮2\)

=> a+b+c+d là hợp số

Ta có: \(a^2+b^2+c^2>2\left(a+b+c\right)\)

\(\Leftrightarrow\) \(a^2+b^2+c^2+3>2\left(a+b+c\right)\) (Vì 3>0)

\(\Leftrightarrow\) \(a^2+b^2+c^2-2a-2b-2c+3>0\)

\(\Leftrightarrow\) \(\left(a^2-2a+1\right)\)+\(\left(b^2-2b+1\right)\)+\(\left(c^2-2c+1\right)\) \(>0\)

\(\Leftrightarrow\) \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2>0\) (luôn đúng \(\forall a,b,c\))

Vậy \(\forall a,b,c\) thì \(a^2+b^2+c^2>2\left(a+b+c\right)\)

Ta có: \(a^2\) \(+\) \(b^2\)

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

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

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

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

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

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+\left(a+b\right)ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=> a = - b hoặc b = - c hoặc c = - a (đpcm)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

 Sửa đề \(VT=\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\)(1)

\(VP=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2.k^2+d^2.k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)(2)

Từ (1) ( 2) => VT = VP (ĐPCM)

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

\(\Leftrightarrow a^2+b^2+c^2+d^2-a\left(b+c+d\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-ab-ac-ad=0\)

\(\Leftrightarrow4\left(a^2+b^2+c^2+d^2-ab-ac-ad\right)=0\)

\(\Leftrightarrow a^2-4ab+4b^2+a^2-4ac+4c^2+a^2-4ad+4d^2+a^2=0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+a^2=0\)

Dấu " = " xảy ra khi : a = 2b = 2c = 2d = 0 <=> a = b = c = d = 0

A²=b.(a-c)-c.(a-b)

A²= ba - bc - ca + cb

A² = ( ba - ca ) + ( bc - cb ) 

A² = a. ( b - c ) + 0

Với a = -20 , b-c = -5  thì:

A² = a. ( b - c ) 

A² = -20 . ( - 5 )

A² = 100

Ta có : 100 = 10 . 10

\(\Rightarrow\)A = 10.

Vậy A = 10

~ HOK TỐT ~

Có b - c = ( - 5 )<=>\(b=c-5\)

Thay \(a=-20\),\(b=c-5\)vào \(A\)ta có

\(A^2=\)_{\(\left(c-5\right)\left(-20-c\right)-c\left(-20-c+5\right)\)}

     _{\(=-20c-c^2+100+5c-c\left(-15-c\right)\)}

   _{\(=100-15c-c^2+15c+c^2\)\(=100\)}

\(\Rightarrow A=10\)hoặc \(A=-10\)