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

Dấu '=' xảy ra khi x=-1

2: \(=-\left(4x^2-12x-10\right)\)

\(=-\left(4x^2-12x+9-19\right)\)

\(=-\left(2x-3\right)^2+19< =19\)

Dấu '=' xảy ra khi x=3/2

3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)

Dấu '=' xảy ra khi x=-2

1 tháng 8 2020

Bài 1 :

a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)

\(=15x^3-6x^2-3x\)

b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)

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

\(=-x^3y+2x^2y^2-3xy\)

c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)

\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)

\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)

d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)

\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)

\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)

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

\(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)

\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)

1 tháng 8 2020

Bài 2 :

3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15

Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)

\(=-\frac{15}{2}-3+15=\frac{9}{2}\)

b) 25x - 4(3x - 1) + 7(5 - 2x)

= 25x - 12x + 4  + 35 - 14x

= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39

Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37

c) 4x - 2(10x + 1) + 8(x - 2)

= 4x - 20x - 2 + 8x - 16

= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18

Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)

d) Tương tự

Bài 3:

a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)

=> 2x2 - 8x - 2x2 - 3x = 4

=> (2x2 - 2x2) + (-8x - 3x) = 4

=> -11x = 4

=> x = \(-\frac{4}{11}\)

b) x(5 - 2x) + 2x(x - 7) = 18

=> 5x - 2x2 + 2x2 - 14x = 18

=> 5x - 14x = 18

=> -9x = 18

=> x = -2

Còn 2 câu làm tương tự

20 tháng 6 2018

Những hằng đẳng thức đáng nhớ

20 tháng 6 2018

Giải:

5) \(-x^2+x-\dfrac{1}{2}\)

\(=-x^2+x-\dfrac{1}{4}+\dfrac{3}{4}\)

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

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

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

6) \(-\dfrac{1}{4}x^2+x-2\)

\(=-\dfrac{1}{4}x^2+x-1-1\)

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

\(=-\left(\dfrac{1}{2}x-1\right)^2-1\le-1\)

\(\Leftrightarrow\dfrac{1}{2}x-1=0\Leftrightarrow x=2\)

Vậy ...

7) \(-\dfrac{1}{9}x^2-\dfrac{1}{3}x+1\)

\(=-\dfrac{1}{9}x^2-\dfrac{1}{3}x-\dfrac{1}{4}+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{9}x^2+\dfrac{1}{3}x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{3}x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

8) \(-2x^2+2xy-2y^2+2x+2y-8\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)

Vậy ...

5 tháng 10 2020

a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3

b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81

c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3

d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2

e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2

= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )

= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6

= -3x2 + 39x + 6

= -3( x2 - 13x - 2 )

f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3

= x3 + y3 + x3 - y3 - 2x3

= 0

g) x2 + 2x( y + 1 ) + y2 + 2y + 1

= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )

= x2 + 2x( y + 1 ) + ( y + 1 )2

= ( x + y + 1 )2

= [ ( x + y ) + 1 ]2

= ( x + y )2 + 2( x + y ) + 1

= x2 + 2xy + y2 + 2x + 2y + 1

1.(x+1)(x-7)+17=(x-3)2+1>0

2.-20-(x-5)(x+3)=-34-(x-1)2<0

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

4.x2+y2+2x+2y+3=(x+1)2+(y+1)2+1>0

5.2x2+2x+y2+2y+5=2(x+1/2)2+(y+1)2+2>0

6.2x2+2y2+2xy+2x+4y+6=(x+y)2+(x+1)2+(y+2)2+1>0

7.-y2+4y-4-/x+1/=-(y-2)2-/x+1/≤0

18 tháng 6 2018

Giải:

1) \(9x^2-12xy+4y^2-3\)

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

\(=\left(3x-2y-\sqrt{3}\right)\left(3x-2y+\sqrt{3}\right)\) (Bước này chắc không cần)

2) \(x^2+4x+1\)

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

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

\(=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)\)

(Bước này chắc không cần)

3) \(x^2-4x+7\)

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

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

4) \(x^2+6x+15\)

\(=x^2+6x+9+6\)

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

5) \(x^2-x+\dfrac{1}{3}\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)

6) \(\dfrac{1}{4}x^2+x\)

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

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

7) \(3x^2+2x+1\)

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

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

8) \(2x^2-2x+1\)

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

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

9) \(10a^2+5b^2+12ab+4a-6b+15\)

\(=4a^2+6a^2+4b^2+b^2+12ab+4a-6b+15\)

\(=\left(6a^2+b^2+12ab\right)+4a+4a^2-6b+4b^2+15\)

\(=\left(6a+b\right)^2+4a\left(1+a\right)-2b\left(3+2b\right)+15\)

18 tháng 6 2018

Giải:

1) \(9x^2-12xy+4y^2-3\)

\(=\left(9x^2-12xy+4y^2\right)-3\)

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

2) \(x^2+4x+1\)

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

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

3) \(x^2-4x+7\)

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

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

4) \(x^2+6x+15\)

\(=x^2+6x+9+6\)

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

5) \(x^2-x+\dfrac{1}{3}\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)

6) \(\dfrac{1}{4}x^2+x\)

\(=x\left(\dfrac{1}{4}x+1\right)\)

7) \(3x^2+2x+1\)

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

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

8) \(2x^2-2x+1\)

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

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

9) \(10a^2+5b^2+12ab+4a-6b+15\)

\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)

\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)

\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)

Vậy ...