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21 tháng 9 2020

a) ( x - 3 )2 - 4 = 0

<=> ( x - 3 )2 - 22 = 0

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

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22

<=> 4x2 + 12x + 9 - 4x2 + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16

<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16

<=> 16x2 - 9 - 16x2 + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x3 - 9x2 + 27x - 27 = -8

<=> ( x - 3 )3 = -8

<=> ( x - 3 )3 = (-2)3

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )3 - x2( x + 3 ) = 2

<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

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

<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5

<=> x3 + 12x - 8 - x3 + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

21 tháng 9 2020

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

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

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

21 tháng 9 2020

a) 2( x - 1 )2 - 4( 3 + x )2 + 2x( x - 5 )

= 2( x2 - 2x + 1 ) - 4( 9 + 6x + x2 ) + 2x2 - 10x

= 2x2 - 4x + 2 - 36 - 24x - 4x2 + 2x2 - 10x

= ( 2x2 - 4x2 + 2x2 ) + ( -4x - 24x - 10x ) + ( 2 - 36 )

= -38x - 34

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

= 2( 4x2 + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )

= 8x2 + 40x + 50 + 3( 16x2 - 1 )

= 8x2 + 40x + 50 + 48x2 - 3

= 56x2 + 40x + 47

c) ( x - 1 )3 - x( x - 3 )2 + 1

= x3 - 3x2 + 3x - 1 - x( x2 - 6x + 9 ) + 1

= x3 - 3x2 + 3x - x3 + 6x2 - 9x

= 3x2 - 6x

d) ( x + 2 )3 - x2( x + 6 ) 

= x3 + 6x2 + 12x + 8 - x3 - 6x2

= 12x + 8

e) ( x - 2 )( x + 2 ) - ( x + 1 )3 - 2x( x - 1 )2

= x2 - 4 - ( x3 + 3x2 + 3x + 1 ) - 2x( x2 - 2x + 1 )

= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x

= -3x3 + 2x2 - 5x - 5 

f) ( a + b - c )2 - ( b - c )2 - 2a( b - c )

= [ ( a + b ) - c ]2 - ( b2 - 2bc + c2 ) - 2ab + 2ac

= [ ( a + b )2 - 2( a + b )c + c2 ] - b2 + 2bc - c2 - 2ab + 2ac

= a2 + 2ab + b2 - 2ac - 2bc + c2 - b2 + 2bc - c2 - 2ab + 2ac

= a2

21 tháng 9 2020

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

Dùng hẳng đẳng thức thứ nhất + hai :

\(2\left(x^2-2\cdot x\cdot1+1^2\right)-4\left(3^2+2\cdot3\cdot x+x^2\right)+2x^2-10x\)

\(2\left(x^2-2x+1\right)-4\left(9+6x+x^2\right)+2x^2-10x\)

\(2x^2-4x+2-36-24x-4x^2+2x^2-10x\)

\(-38x-34\)

b) 2(2x + 5)2 - 3(4x + 1)(1 - 4x)

Dùng đẳng thức thứ 1 + 3

= 2[(2x)2 + 2.2x.5 + 52 ] - (-3)[(4x)2 - 12 ]

= 2(4x2 + 20x + 25) - (-3).(16x2 - 1)

= 8x2 + 40x + 50 - (3 - 48x2)

= 8x2 + 40x + 50 - 3 + 48x2

= 56x2 + 40x + 47

c) (x - 1)3 - x(x - 3)2 + 1

Dùng đẳng thức 2 + 5:

= x3 - 3.x2.1 + 3.x.12 - 13 - x(x2 - 2.x.3 + 32) + 1

= x3 - 3x2 + 3x - 1 - x3 + 6x2 - 9x + 1

= (x3 - x3) + (-3x2 + 6x2) + (3x - 9x) + (-1 + 1)

= 3x2 - 6x

d) (x + 2)3 - x2(x + 6)

= x3 + 3.x2.2 + 3.x.22 + 23 - x3 - 6x2

= x3 + 6x2 + 12x + 8 - x3 - 6x2

= (x3 - x3) + (6x2 - 6x2) + 12x + 8 = 12x + 8

e) Dùng đẳng thức thứ 3,4 và 2

= x2 - 4 - (x3 + 3.x2.1 + 3.x.12 + 13) - 2x(x2 - 2.x.1 + 12)

= x2 - 4 - (x3 + 3x2 + 3x + 1) - 2x3 + 4x2 - 2x

= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x

= (x2 - 3x2 + 4x2) + (-4 - 1) + (-x3 - 2x3) + (-3x - 2x)

= 2x2 - 5 - 3x3 - 5x

f) Đặt \(a+b-c=A\)

\(b-c=B\)

\(A^2-B^2-2AB\)

\(A^2-2AB+\left(-B\right)^2\)

\(=A^2-2AB+B^2\)

= (A - B)2

= (a + b - c - (b - c))2

= (a + b - c - b + c)2

= a2

2 tháng 8 2018

(x + 2)(x - 2) - (x - 2)(x + 5)

= (x - 2)(x + 2 - x - 5)

= (x - 2)-3

= -3x + 6

b) 2x(3x2y + 4x2y - 3)

= 2x(7x2y - 3)

= 14x3y - 6x

2 tháng 8 2018

bạn giải hết đc ko ạ

9 tháng 8 2021

1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

9 tháng 8 2021

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

10 tháng 9 2020

a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

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

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

25 tháng 12 2020

ko có biết

2 tháng 6 2018

1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)

2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)

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

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

5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)

\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)

\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)

7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)

\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)

\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

2 tháng 6 2018

|2 - x|2 + 6x - 3 = 0

<=> (x - 2)2 + 6x - 3 = 0

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

<=> x2 + 2x + 1 = 0

<=> (x + 1)2 = 0

<=> x + 1 = 0

<=> x = -1

Bắt phải thể hiện -_-

5 tháng 10 2020

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

5 tháng 10 2020

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

13 tháng 7 2021

bài vách ngọc ngà và bài cà phê ko đường