( x - 1 )³ = ( 2x + 1 )³ + 3( x + 2 )

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

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

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

<=> ( 2x + 1 - x + 1 )( 4x² + 4x + 1 + 2x² - x - 1 + x² - 2x + 1 ) + 3( x + 2 ) = 0

<=> ( x + 2 )( 7x² + x + 1 ) + 3( x + 2 ) = 0

<=> ( x + 2 )( 7x² + x + 1 + 3 ) = 0

<=> ( x + 2 )( 7x² + x + 4 ) = 0

Vì 7x² + x + 4 = 7( x² + 1/7x + 1/196 ) + 111/28 = 7( x + 1/14 )² + 111/28 ≥ 111/28 > 0 ∀ x

=> x + 2 = 0 => x = -2

M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)

= a³ + b³ + 3a³b + 3ab³ + 6a²b²

= (a + b)³ - 3ab(a + b) + 3a³b + 3ab³ + 6a²b²

= 1³ + 3ab[-(a + b) + a² + b² + 2ab]

= 1 - 3ab[-1+ (a + b)²]

= 1 - 3ab(-1 + 1)

= 1 - 3ab.0 = 1

M = a³ + b³ + 3ab( a² + b² ) + 6a²b²( a + b )

= ( a + b )³ - 3ab( a + b ) + 3ab( a² + b² ) + 6a²b²

= 1 - 3ab + 3a³b + 3ab³ + 6a²b²

= 1 - 3ab( 1 - a² - b² - 2ab )

= 1 - 3ab[ 1 - ( a² + 2ab + b² ) ]

= 1 - 3ab[ 1 - ( a + b )² ]

= 1 - 3ab( 1 - 1 )

= 1 - 3ab.0 = 1

a) 4x⁴ + 4x³ + 5x² + 2x + 1 

= 4x⁴ + 2x³ + 2x³ + 2x² + 2x² + x² + x + x + 1

= ( 4x⁴ + 2x³ + 2x² ) + ( 2x³ + x² + x ) + ( 2x² + x + 1 )

= 2x²( 2x² + x + 1 ) + x( 2x² + x + 1 ) + 1( 2x² + x + 1 )

= ( 2x² + x + 1 )( 2x² + x + 1 )

= ( 2x² + x + 1 )²

b) x⁴ - 7x³ + 14x² - 7x + 1 

Sau khi phân tích thì đa thức có dạng ( x² + ax + 1 )( x² + bx + 1 )

x⁴ - 7x³ + 14x² - 7x + 1 = ( x² + ax + 1 )( x² + bx + 1 )

<=> x⁴ - 7x³ + 14x² - 7x + 1 = x⁴ + bx³ + x² + ax³ + abx² + ax + x² + bx + 1

<=> x⁴ - 7x³ + 14x² - 7x + 1 = x⁴ + ( a + b )x³ + ( ab + 2 )x² + ( a + b )x + 1

Đồng nhất hệ số ta có :

\(\hept{\begin{cases}a+b=-7\\ab+2=14\end{cases}}\)=> a = -4 ; b = -3 hoặc a = -3 ; b = -4 ( giải cái này bạn có thể lên coccoc )

=> x⁴ - 7x³ + 14x² - 7x + 1 = ( x² - 4x + 1 )( x² - 3x + 1 ) 

CẢM ƠN BẠN NHÁ

Ta có: \(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(2x+2y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}\)

Thay vào biểu thức A, ta có:

\(A=\left(1-1\right)^{2014}+\left(1-2\right)^{2015}+\left(-1+1\right)^{2016}\)

\(=0-1+0=-1\)

Vậy GTBT A=-1 tại x=1, y=-1

chia hết rồi sao lại dư :))? tớ sửa lại đề nhé ! có đúng ko thì ko bt :P

\(4x^2+ax+b⋮x-2\)

Để \(4x^2+ax+b⋮x-2\)

<=> \(x\left(a+b\right)-x+2+b=0\)

<=> \(ax+bx-x+2+b=0\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+7x+6x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+7\right)+6\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+6\right)+\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)\(\Leftrightarrow\frac{2x+10}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+7\right)+\left(x+4\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3x+18}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3\left(x+6\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x+30.25=56.25\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+5.5\right)^2=\left(7.5\right)^2\\\left(x+5.5\right)^2=\left(-7.5\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+5.5=7.5\\x+5.5=-7.5\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=7.5-5.5=2\\x=-7.5-5.5=-13\end{cases}}\)

\(\text{Vậy x }\in\left\{2;-13\right\}\)

\(\text{nhớ tích cho mk nha Thanh Do}\)

\(x-x^2-2\)

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

Ta có : \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge0\)

mà \(x\le x^2\)

\(\Rightarrow x< x^2+2\)

\(\Rightarrow x-\left(x^2+2\right)< 0\)

\(\Rightarrow x-x^2-2< 0\)

Ta có: \(x-x^2-2=x-x^2-\frac{1}{4}-\frac{7}{4}\)

\(=\left(x-x^2-\frac{1}{4}\right)-\frac{7}{4}=-\left(x^2-x+\frac{1}{4}\right)-\frac{7}{4}\)

\(=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)-\frac{7}{4}=-\left(x-\frac{1}{2}\right)^2-\frac{7}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\frac{1}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)\(\forall x\)

\(\Rightarrow x-x^2-2\le-\frac{7}{4}\)\(\forall x\)

hay \(x-x^2-2< 0\)( đpcm ) 

Bài làm 

Xét tam giác MNQ ta có : 

E là trung điểm MN 

H là trung điểm MQ

=)) EH là đường TB tam giác MNQ

=)) EH // QN và EH = 1/2 QN (1)

Xét tam giác PNQ ta có : 

F là trung điểm MP 

G là trung điểm QP 

=)) FG là đường TB tam giác PNQ

=)) FG // NQ và FE = 1/2 NQ (2)

Từ 1 ; 2 =)) tứ giác EFGH là hình bình hành