Đặt \(A=\frac{5}{x^2+2x+5}\)

Để A đạt GTLN => x² + 2x + 5 đạt GTNN

Ta có : x² + 2x + 5 = ( x² + 2x + 1 ) + 4 = ( x + 1 )² + 4 ≥ 4 ∀ x

=> GTNN của x² + 2x + 5 = 4 khi x = -1

=> MinA = 5/4 <=> x = -1

3x( x - 1 ) + x - 1 = 0

⇔ 3x( x - 1 ) + ( x - 1 ) = 0

⇔ ( x - 1 )( 3x + 1 ) = 0

⇔ x - 1 = 0 hoặc 3x + 1 = 0

⇔ x = 1 hoặc x = -1/3

( 3x - 4 )( x - 2 ) = 3x( x - 9 ) - 3

⇔ 3x² - 10x + 8 = 3x² - 27x - 3

⇔ 3x² - 10x - 3x² + 27x = -3 - 8

⇔ 17x = -11

⇔ x = -11/17

Trả lời :

(3x - 4)(x - 2) = 3x(x -9) - 3

=> 3x² - 10x + 8 - 3x² + 27x + 3 = 0

=> 17x + 11 = 0

=> 17x = - 11

=> \(x=\frac{-11}{17}\)

a² + b² + 2ab + 2a + 2b + 1

= ( a² + 2ab + b² ) + ( 2a + 2b ) + 1

= ( a + b )² + 2( a + b ) + 1²

= ( a + b + 1 )²

3x( x - 2y ) - 6y( 2y - x )

= 3x( x - 2y ) + 6y( x - 2y )

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

x² + 2x - 3

= x² - x + 3x - 3

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

= ( x - 1 )( x + 3 )

a) \(a^2+b^2+2ab+2a+2b+1\)

\(=\left(a^2+2ab+b^2\right)+\left(2a+2b\right)+1\)

\(=\left(a+b\right)^2+2\left(a+b\right)+1\)

\(=\left(a+b+1\right)^2\)

b) \(3x\left(x-2y\right)-6y\left(2y-x\right)\)

\(=3x\left(x-2y\right)+6y\left(x-2y\right)\)

\(=3\left(x-2y\right)\left(x+2y\right)\)

c) \(x^2+2x-3=x^2-x+3x-3\)

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

\(=x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x+3\right)\)

\(x^3+4x^2+8x=-5\)

\(\Leftrightarrow x^3+4x^2+8x+5=0\)

\(\Leftrightarrow x^3+x^2+3x^2+3x+5x+5=0\)

\(\Leftrightarrow\left(x^3+x^2\right)+\left(3x^2+3x\right)+\left(5x+5\right)=0\)

\(\Leftrightarrow x^2.\left(x+1\right)+3x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+3x+5\right)=0\)(1)

Ta có: \(x^2+3x+5=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\forall x\)

\(\Rightarrow x^2+3x+5\ge\frac{11}{4}\)(2)

Từ (1) và (2) \(\Rightarrow x+1=0\)\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)