( 3x - 5 )( x + 2 ) = x² - 5x 

<=> 3x² + 6x - 5x - 10 - x² + 5x = 0

<=> 2x² + 6x - 10 = 0

Δ = b² - 4ac = 6² - 4.2.(-10) = 36 + 80 = 116

Δ > 0 nên phương trình có hai nghiệm phân biệt :

\(x_1=\frac{-b+\sqrt{\text{Δ}}}{2a}=\frac{-6+\sqrt{116}}{4}=\frac{-3+\sqrt{29}}{2}\)

\(x_2=\frac{-b-\sqrt{\text{Δ}}}{2a}=\frac{-6-\sqrt{116}}{4}=\frac{-3-\sqrt{29}}{2}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{-3\pm\sqrt{29}}{2}\right\}\)

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

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

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

\(\Leftrightarrow2x^2+6x-10=0\)

\(\Leftrightarrow2\left(x^2+3x-5\right)=0\Leftrightarrow x^2+3x+5=0\)giải delta ta được : 

\(x=\frac{-3\pm\sqrt{29}}{2}\)

ta có 

\(PT\Leftrightarrow\left(x-1\right)\left(2x+3\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

( x - 1 )( 2x + 3 ) + 2x = 2

<=> ( x - 1 )( 2x + 3 ) + 2x - 2 = 0

<=> ( x - 1 )( 2x + 3 ) + 2( x - 1 ) = 0

<=> ( x - 1 )( 2x + 3 + 2 ) = 0

<=> ( x - 1 )( 2x + 5 ) = 0

<=> x - 1 = 0 hoặc 2x + 5 = 0

<=> x = 1 hoặc x = -5/2

Vậy tập nghiệm của phương trình là : S = { 1 ; -5/2 }

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

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

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

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

\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)

\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)

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

2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)

Thay x = 14 ; y = -15 vào biểu thức ta được 

\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)

Tìm GTNN??

Ta có: \(A=\frac{2x^2+2x+7}{x^2+x+1}=\frac{2\left(x^2+x+1\right)+5}{x^2+x+1}=2+\frac{5}{x^2+x+1}\)

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

\(\Rightarrow A=2+\frac{5}{x^2+x+1}\le2+\frac{5}{\frac{3}{4}}=\frac{26}{3}\)

Dấu "=" xảy ra khi: x = -1/2

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

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

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

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

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

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

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

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

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

\(=-2^2\)

\(=-4\)