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Bài 1: 

b: \(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)

\(=8x^2+40x+50+48x^2-3\)

\(=56x^2+40x+47\)

Bài 2: 

b: \(\Leftrightarrow3x-3+9x-18=2x-6+4x-4\)

=>12x-21=6x-10

=>6x=11

hay x=11/6

28 tháng 8 2021

a. y4 - 14y2 + 49

Gọi y2 là t, ta có:

t2 - 14t + 49

<=> t2 - 14t + 72

<=> (t - 7)2

Thay x2 = t

<=> (x2 - 7)2

b. \(\dfrac{1}{4}-x^2\)

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

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

c. x4 - 16

<=> (x2)2 - 42

<=> (x2 - 4)(x2 + 4)

d. x2 - 9

<=> x2 - 32

<=> (x - 3)(x + 3)

 

27 tháng 6 2021

a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)

\(=-18x^3-46x^2-8x+16\).

 

a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)

\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)

\(=-5x^3-38x^2-76x-46\)

b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)

\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)

\(=x^3-20x^2+56x+49\)

c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)

\(=-18x^3+2x^2+40x+16\)

22 tháng 10 2023

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

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

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

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

21 tháng 7 2021

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

5 tháng 7 2016

Các bạn cố gắng giúp mình nha . Mình xin chân thành cảm ơn 

20 tháng 7 2018

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x2 - 10x - 2x2 + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5

<=> 2x+ x + 6x + 3 - 2x2 - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

19 tháng 7 2018

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

7 tháng 12 2019

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)

14 tháng 7 2019

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha , 

20 tháng 10 2020

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )2

= 6x2 - 4x - ( x2 + 4x + 4 )

= 6x2 - 4x - x2 - 4x - 4

= 5x2 - 8x - 4

b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x3 + 8 + 2( x + 1 )( x - 1 )

= x3 + 8 + 2( x2 - 1 )

= x3 + 8 + 2x2 - 2

= x3 + 2x2 + 6

c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2

= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1

= 4

d) x2 - 3x + xy - 3y

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

= ( x - 3 )( x + y )

Bài 2.

a) 4x2 - 4xy + y2 = ( 2x - y )2

b) 9x3 - 9x2y - 4x + 4y

= 9x2( x - y ) - 4( x - y )

= ( x - y )( 9x2 - 4 )

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

c) x3 + 2 + 3( x3 - 2 )

= x3 + 2 + 3x3 - 6

= 4x3 - 4

= 4( x3 - 1 )

= 4( x - 1 )( x2 + x + 1 )

Bài 3.

2( x - 2 ) = x2 - 4x + 4

⇔ ( x - 2 )2 - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4