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19 tháng 7 2019

a) \(x^3-16x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{16}=\pm4\end{cases}}\)

Vậy \(x\in\left\{0;\pm4\right\}\)

19 tháng 7 2019

b) \(x^2-6x+9=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

8 tháng 8 2016

d) <=>x2-5x-x+5=0

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

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

<=>x=5 hoặc x=1

9 tháng 8 2016

thank nha

3 tháng 3 2020

a,

đoạn 9x-6-> 2x-6=0

=> x=3

b,6x^2+13x+5=6x^2-20x+6

33x=1

=>x=1/33

3 tháng 3 2020

a) (x+1)(x+9)=(x+3)(x+5) 

<=>x^2+10x+9=x^2+8x+15

<=>x^2+10x+9-x^2-8x-15=0

<=>9x-6=0 phải là 2x - 6

<=>9x=6

<=>x=6/9=2/3 => S= 2/3

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

<=>6x^2+13x+5=6x^2-16x+6 phải là 6x^2 - 20x + 6

<=>6x^2+13x+5-6x^2+16x-6=0

<=>29x-1=0

<=>29x=1

<=>x=1/29

13 tháng 9 2020

a) x4 - 16x2 = 0

<=> ( x2 )2 - ( 4x )2 = 0

<=> ( x2 - 4x )( x2 + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x2( x - 4 )( x + 4 ) = 0

<=> \(\hept{\begin{cases}x^2=0\\x-4=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé )

b) 9x2 + 6x + 1 = 0

<=> ( 3x )2 + 2.3x.1 + 12 = 0

<=> ( 3x + 1 )2 = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x2 - 6x = 16

<=> x2 - 6x - 16 = 0

<=> x2 + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x2 + 6x = 80

<=> 9x2 + 6x - 80 = 0

<=> 9x2 + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) Áp dụng công thức an.bn = ( ab )n ta có :

25( 2x - 1 )2 - 9( x + 1 )2 = 0

<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0 

<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0

<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

<=> \(\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

16 tháng 9 2020

             Bài làm :

a) x4 - 16x2 = 0

<=> ( x2 )2 - ( 4x )2 = 0

<=> ( x2 - 4x )( x2 + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x2( x - 4 )( x + 4 ) = 0

 Vậy x=0 hoặc x=±4

b) 9x2 + 6x + 1 = 0

<=> ( 3x )2 + 2.3x.1 + 12 = 0

<=> ( 3x + 1 )2 = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x2 - 6x = 16

<=> x2 - 6x - 16 = 0

<=> x2 + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x2 + 6x = 80

<=> 9x2 + 6x - 80 = 0

<=> 9x2 + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) 25( 2x - 1 )2 - 9( x + 1 )2 = 0

<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0 

<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0

<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

13 tháng 9 2020

a) Ta có : x4 - 16x2 = 0

=> x4 - 8x2 - 8x2 + 64 = 64

=> x2(x2 - 8) - 8(x2 - 8) = 64

=> (x2 - 8)2 = 64

=> \(\orbr{\begin{cases}x^2-8=8\\x^2-8=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=16\\x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm4\\x=0\end{cases}}\Rightarrow x\in\left\{4;-4;0\right\}\)

b) Ta có 9x2 + 6x + 1 = 0

=> 9x2 + 3x + 3x + 1 = 0

=> 3x(3x + 1) + (3x + 1) = 0

=> (3x + 1)2 = 0

=> 3x + 1 = 0

=> x = -1/3

c) Ta có x2 - 6x = 16

=> x2 - 6x + 9 = 25

=> (x - 3)2 = 25

=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\Rightarrow x\in\left\{8;-2\right\}\)

d) 9x2 + 6x = 80

=> 9x2 + 6x + 1 = 81

=> (3x + 1)2 = 81

=> \(\orbr{\begin{cases}3x+1=9\\3x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\Rightarrow x\in}\left\{\frac{8}{3};\frac{-10}{3}\right\}\)

e) 25(2x - 1)2 - 9(x + 1)2 = 0

=> [5(2x - 1)]2 - [3(x + 1)]2 = 0

=> (10x - 5)2 - (3x + 3)2 = 0

=> (10x - 5 - 3x - 3)(10x - 5 + 3x + 3) = 0

=> (7x - 8)(13x - 2) = 0

=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

27 tháng 9 2021

ko bt lm:)

31 tháng 10 2019

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

\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

31 tháng 10 2019

\(B1:\)

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

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

19 tháng 7 2016

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

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)

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

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

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

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

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

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

c)\(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

e)\(x^2-2x-3=0\)

\(\Leftrightarrow x^2+x-3x-3=0\)

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

19 tháng 7 2016

Cảm ơn bạn nha

a) Ta có: \(x^4-16x^2=0\)

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

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

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

Vậy: \(x\in\left\{0;4;-4\right\}\)

b) Ta có: \(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

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

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\frac{1}{3}\)

Vậy: \(x=-\frac{1}{3}\)

c) Ta có: \(x^2-6x=16\)

\(\Leftrightarrow x^2-6x-16=0\)

\(\Leftrightarrow x^2-8x+2x-16=0\)

\(\Leftrightarrow x\left(x-8\right)+2\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-2\right\}\)

d) Ta có: \(9x^2+6x=80\)

\(\Leftrightarrow9x^2+6x-80=0\)

\(\Leftrightarrow9x^2+6x+1-81=0\)

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

\(\Leftrightarrow\left(3x+1-9\right)\left(3x+1+9\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=8\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)

e) Ta có: \(25\left(2x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(10x-5\right)^2-\left(3x+3\right)^2=0\)

\(\Leftrightarrow\left(10x-5-3x-3\right)\left(10x-5+3x+3\right)=0\)

\(\Leftrightarrow\left(7x-8\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-8=0\\13x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=8\\13x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{7}\\x=\frac{2}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{7};\frac{2}{13}\right\}\)