Bài 1:

\(6x^2-2\left(x-y\right)^2-6y^2\)

\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)

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

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

Bài 2:

\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)

\(=\left(3x-1-x-1\right)^2\)

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

b) Thay \(x=\frac{9}{4}\)vào (1) ta được: 

\(\left(2.\frac{9}{4}-2\right)^2\)

\(=\frac{25}{4}\)

Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)

Bài 3:

Ta có: \(M=x^2+4x+5\)

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

Vì \(\left(x+2\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)

Hay \(M\ge1;\forall x\)

Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)

                       \(\Leftrightarrow x=-2\)

Vậy \(M_{min}=1\Leftrightarrow x=-2\)

Bài 1 : trên là sai nha mình làm lại

\(6x^2-2\left(x-y\right)^2-6y^2\)

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

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

\(=2\left(x-y\right)\left(2x+4y\right)\)

\(=4\left(x-y\right)\left(x+2y\right)\)

tách nhỏ câu hỏi ra bạn

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

các bạn làm giùm mih đi câu nào cũng được

mk mới lớp 5 nên ko bt

Câu 1:

a) \(2x^2+5x-3=\left(2x^2+6x\right)-\left(x+3\right)\)

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

b) \(x^4+2009x^2+2008x+2009\)

\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)

c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-5-\sqrt{5}\\x=-5+\sqrt{5}\end{cases}}\)

Câu 1.

a) 2x² + 5x - 3 = 2x² + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )

b) x⁴ + 2009x² + 2008x + 2009 

= x⁴ + 2009x² + 2009x - x + 2009 

= ( x⁴ - x ) + ( 2009x² + 2009x + 2009 )

= x( x³ - 1 ) + 2009( x² + x + 1 )

= x( x - 1 )( x² + x + 1 ) + 2009( x² + x + 1 )

= ( x² + x + 1 )[ x( x - 1 ) + 2009 ]

= ( x² + x + 1 )( x² - x + 2009 )

c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )

Câu 2. 

3x² + x - 6 - √2 = 0

<=> ( 3x² - 6 ) + ( x - √2 ) = 0

<=> 3( x² - 2 ) + ( x - √2 ) = 0

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

<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0

<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)

+) x - √2 = 0 => x = √2

+) 3( x + √2 ) + 1 = 0

<=> 3( x + √2 ) = -1

<=> x + √2 = -1/3

<=> x = -1/3 - √2

Vậy S = { √2 ; -1/3 - √2 }

Câu 3.

A = x( x + 1 )( x² + x - 4 )

= ( x² + x )( x² + x - 4 )

Đặt t = x² + x

A = t( t - 4 ) = t² - 4t = ( t² - 4t + 4 ) - 4 = ( t - 2 )² - 4 ≥ -4 ∀ t

Dấu "=" xảy ra khi t = 2

=> x² + x = 2

=> x² + x - 2 = 0

=> x² - x + 2x - 2 = 0

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

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

=> x = 1 hoặc x = -2

=> MinA = -4 <=> x = 1 hoặc x = -2

\(x^2-\left(y-3\right)^2-4x+4\)

\(=x^2-\left(y^2-6y+9\right)-4x+4\)

\(=x^2-y^2+6y-9-4x+4\)

\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)

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

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

\(=\left(x-y+5\right)\left(x+y-5\right)\)

1.

x² - ( y - 3 )² - 4x + 4

= ( x² - 4x + 4 ) - ( y - 3 )²

= ( x - 2 )² - ( y - 3 )²

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

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

= ( x - y + 1 )( x + y - 5 )

2.

a) Ta có : 2x⁴ + 8x³ + 9x² - 4x - 5

= 2x⁴ + 10x² - x² + 8x³ - 4x - 5

= ( 2x⁴ - x² ) + ( 8x³ - 4x ) + ( 10x² - 5 )

= x²( 2x² - 1 ) + 4x( 2x² - 1 ) + 5( 2x² - 1 )

= ( 2x² - 1 )( x² + 4x + 5 )

=>(2x⁴ + 8x³ + 9x² - 4x - 5) : ( 2x² - 1 ) = x² + 4x + 5

b) Ta có : x² + 4x + 5 = ( x² + 4x + 4 ) + 1 = ( x + 2 )² + 1 ≥ 1 > 0 ∀ x

=> đpcm