\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{n+1-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\)

\(2A=\frac{1}{2}-\frac{1}{n\left(n+1\right)}\)

\(A=\frac{1}{4}-\frac{1}{2n\left(n+1\right)}\)

a³ + b³ + c³ = 3abc 

⇒ a³ + b³ + c³ - 3abc = 0

⇒ ( a³ + b³ ) + c³ - 3abc = 0

⇒ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇒ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇒ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇒ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

Vì a + b + c ≠ 0

⇒ a² + b² + c² - ab - bc - ac = 0

⇒ 2( a² + b² + c² - ab - bc - ac ) = 0

⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

⇒ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

⇒ ( a - b )² + ( b - c )² + ( a - c )² = 0

Vì \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(a-c\right)^2\end{cases}}\ge0\forall a,b,c\)⇒ ( a - b )² + ( b - c )² + ( a - c )² ≥ 0 ∀ a,b,c

Dấu "=" xảy ra khi a = b = c

Khi đó \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Từ \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right).c\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right)c-3ab\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)=0\)

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

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

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)

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

Vì \(a+b+c\ne0\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)\(\forall a,b,c\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

Thay \(a=b=c\)vào N ta có: \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy \(N=\frac{1}{3}\)

x² + 5y² + 2x - 6y - 4xy + 2 = 0

⇔ ( x² - 4xy + 4y² + 2x - 2y + 1 ) + ( y² - 2y + 1 ) = 0

⇔ ( x - 2y + 1 )² + ( y - 1 )² = 0

Vì \(\hept{\begin{cases}\left(x-2y+1\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{cases}}\)⇒ ( x - 2y + 1 )² + ( y - 1 )² ≥ 0 ∀ x, y

Dấu "=" xảy ra khi x = y = 1

Khi đó : S = x²⁰²⁰ + ( y - 2 )²⁰²¹ = 1²⁰²⁰ + ( 1 - 2 )²⁰²¹ = 1 - 1 = 0

a) \(x^2\left(x+1\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x^3+x^2-x^3-9\)

\(=x^2-9\)

\(=\left(x-3\right)\left(x+3\right)\)

b) \(2\left(x^2-1\right)-\left(x-1\right)^2-\left(x+1\right)^2\)

\(=2\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+1\right)^2\)

\(=-\left[\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]\)

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

\(=-2^2\)

\(=-4\)

thôi m ko cần nữa

a, ĐKXĐ : \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)hoặc \(2x^2-x^3\ne0\)hay \(x\ne\pm2;0\)

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

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

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

\(=\frac{-4x^2-6x+3}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{x-3}=\frac{\left(-4x^2-6x+3\right)\left(-x\right)}{\left(x+2\right)\left(x-3\right)}=\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}\)

b, Ta có : A > 0 hay \(\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}>0\)

\(\Leftrightarrow x\left(4x^2+6x-3\right)>0\)

\(\Leftrightarrow4x^2+6x-3>0\) bạn xem lại bài mình có chỗ nào sai ko nhé !!! 

c, Ta có : \(\left|x-7\right|=4\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)

TH1 : Thay x = 11 vào phân thức trên : ... 

TH2 : Thay x = 3 vào phân thức trên : .... tự làm 

\(\left(3x+5\right)\left(x+2\right)=x^2-5x\)

\(\Leftrightarrow3x^2+6x+5x+10=x^2-5x\)

\(\Leftrightarrow3x^2+11x+10-x^2+5x=0\)

\(\Leftrightarrow2x^2+16x+10=0\)

\(\Leftrightarrow2\left(x^2+8x+5\ne0\right)=0\)

Vậy phương trình vô nghiệm