Ta có

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=0\Rightarrow ab+bc+ac=0.\)

\(A=\frac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^2}\)

Ta có

\(\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3-3\left(abc\right)^2=\)

\(=\left(ab+bc+ac\right)\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2-abbc-bcac-abac\right]=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3\left(abc\right)^2\)

\(\Rightarrow A=\frac{3\left(abc\right)^2}{\left(abc\right)^2}=3\)

Từ \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

Ta có: \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)

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

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

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

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

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

\(=\frac{b}{-2b}+\frac{c}{-2c}+\frac{a}{-2a}=\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)=-\frac{3}{2}\)

Em ko bik ạ

Bài 1:

a ) a.( b² + c² ) + b.( a² + c² ) + c.( a² + b² ) + 2abc

= ab² + ac² + a²b + bc² + a²c + b²c + 2abc

= ( ab² + a²b ) + ( ac² + bc² ) + ( a²c + 2abc + b²c )

= ab.( a + b ) + c².( a + b ) + c.( a² + 2ab + b² )

= ab.( a + b ) + c².( a + b )v + c.( a + b)²

= ( a + b ).[ ( ab + c² + c. ( a + b ) ]

= ( a + b ).( ab + c² + ac + bc )

= ( a + b ).[ ( ab + ac ) + ( c² + bc) ]

= ( a + b ).[ a.( b + c ) + c.( b + c ) ]

= ( a + b ).( b + c ).( a + c )

b) ab.( a + b ) - bc.( b + c ) + ac.( a - c )

= ab.( a + b ) - bc.( b + c ) + ac.[ ( a + b  ) - ( b + c ) ]

= ab.( a + b ) - bc. ( b + c ) + ac.( a + b ) - ac.( b + c )

= ab.( a + b ) + ac.( a + b ) - bc.( b + c ) - ac.( b + c )

= ( a + b ).( ab + ac ) + ( b + c ).( -bc - ac )

= ( a + b ).a.( b + c ) - ( b + c ).c.( a + b )

= ( a + b ).( b + c ).( a - c )

c) ( x² + x )² + 2.( x² + x ) - 3

Đặt x² + x = a

Khi đó đa thức trở thành:

a² + 2a - 3

= a² + 3a - a - 3

= a.( a + 3 ) - ( a + 3 )

= ( a - 1 ).( a - 3 )

\(\Rightarrow\) ( x² + x - 1 ).( x² + x - 3 )

B₂

ab.( a - b ) + bc.( b - c ) + ca.( c - a ) = 0

\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.[ ( a - b ) + ( b - c ) ] = 0

\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.( a - b ) - ca.( b - c ) = 0

\(\Leftrightarrow\)ab.( a - b ) - ca.( a - b ) + bc.( b - c ) - ca.( b - c ) = 0

\(\Leftrightarrow\) ( a - b ).( ab - ca ) + ( b - c ).( bc - ca ) = 0

\(\Leftrightarrow\) ( a - b ).a.( b - c ) - ( b - c ).c.( a - b ) = 0

\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0

\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0

\(\Leftrightarrow\) a = b , b = c , a = c

\(\Rightarrow\) a = b = c

         \(a+b+c=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a+b\right)^2=c^2\\\left(b+c\right)^2=a^2\\\left(c+a\right)^2=b^2\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a^2+b^2-c^2=a^2+b^2-\left(a+b\right)^2=-2ab\\b^2+c^2-a^2=b^2+c^2-\left(b+c\right)^2=-2bc\\c^2+a^2-b^2=c^2+a^2-\left(c+a\right)^2=-2ca\end{cases}}\)

Vậy    \(B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)

P/s:  you tham khảo nha, mk ko biết đúng hay sai

Ta có: \(\frac{ab}{a^2+b^2-c^2}\)

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

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

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

\(=\frac{ab}{-2ab}\)

\(=-\frac{1}{2}\)

Tương tự mà tính

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath

\(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(a+c\right)^2-2ac-b^2}=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(a+b+c\right)\left(b+c-a\right)-2bc}+\dfrac{ac}{\left(a+b+c\right)\left(a+c-b\right)-2ac}=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}.3=-\dfrac{3}{2}\)

Từ ab/(a+b)=bc/(b+c). Nhân chéo suy ra a=c

Chứng minh tương tự suy ra  a=b=c

Thay hết thành a vào M tính ra M=1

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