AA',BB',CC' là gì vậy bạn ?

\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)

Sai đề ở chỗ \(\left(\frac{x^2-16}{x-4}+1\right)\)thành -1

x²( x + 1 )² + 2x² + 2x - 8

= x²( x + 1 )² + ( 2x² + 2x ) - 8

= x²( x + 1 )² + 2x( x + 1 ) - 8 (1)

Đặt a = x( x + 1 ) 

Thế vào (1) ta được :

(1) = a² + 2a - 8

= ( a² + 2a + 1 ) - 9

= ( a + 1 )² - 3² 

= ( a + 1 - 3 )( a + 1 + 3 ) 

= ( a - 2 )( a + 4 )

= [ x( x + 1 ) - 2 ][ x( x + 1 ) + 4 ]

= ( x² + x - 2 )( x² + x + 4 )

= ( x² - x + 2x - 2 )( x² + x + 4 )

= [ x( x - 1 ) + 2( x - 1 ) ]( x² + x + 4 )

= ( x - 1 )( x + 2 )( x² + x + 4 )

https://olm.vn/thanhvien/phuongeieu chẳng hiều gì về toán học à bạn ?

Cosi 2 số : \(ab+\frac{1}{a}\ge2ab\frac{1}{a}=2b\)

\(bc+\frac{1}{b}\ge2bc\frac{1}{b}=2c\)

\(ca+\frac{1}{c}\ge2ca\frac{1}{c}=2a\)

Cộng vế với vế ta được : \(2\left(ab+\frac{1}{a}+bc+\frac{1}{b}+ca+\frac{1}{c}\right)\ge2\left(a+b+c\right)\)

Dấu ''='' xảy ra <=> a = b = c 

*Gỉa sử : a = b = c = 1 ta được : \(A=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=1\)

Ta có :

 \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)

\(\Rightarrow2\left(ab+bc+ca\right)=0\)

\(\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\frac{ab+bc+ca}{abc}=0\)

\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=0\)

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

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

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

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

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab\left(-\frac{1}{c}\right)}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\) (ĐPCM)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

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

\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

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

\(\Rightarrow a+b+c=abc\) (ĐPCM)