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6 tháng 3 2020

đề là gì

a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)

vậy x=2/3 hoặc x=-6

6 tháng 3 2020

a, (3x-2) (x+6) (x^2 +5) = 0 

<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x2 + 5 = 0 (loại vì x2 \(\ge\)0 => x2 + 5 > 0)

<=> x = 2/3 hoặc x = -6 

b, (2x+5)^2 = (3x-1)^2 

<=> (2x + 5)2 - (3x - 1)2 = 0

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)

c, 4x2 (x-1) - x+1 = 0

<=> 4x2(x - 1) - (x - 1) = 0

<=> (x - 1)(4x2 - 1) = 0

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

vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0

hay x = 1 hoặc x = 1/2 hoặc x = -1/2

16 tháng 8 2018

Bài 1:

  a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10

b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61

Bài 2:

a)(2x-1)^2-(x+3)^2 = 0

   <=> (2x-1-x-3).(2x-1+x+3) =0

   <=>(x-4).(3x+2) = 0

<=> x-4 = 0 hoặc 3x+2=0 

              *x-4=0    =>   x=4

              *3x+2 = 0     => 3x=-2   => x=-2/3

b)x^2(x-3)+12-4x=0       <=>     x^2(x-3) - 4(x-3) =0     <=>       (x-3).(x-2)(x+2)   <=> x-3=0 hoặc x-2=0  hoặc x+2 =0

                                                                                        *x-3=0  => x=3

                                                                                        *x-2=0    =>x=2

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

c)  6x^3 -24x =0  <=> 6x(x^2 -4)=0    <=> 6x(x-2)(x+2)=0    <=>  x=0 hoặc x-2 =0 hoặc x+2=0  <=> x=0 hoặc x=2  hoặc x=-2

16 tháng 5 2019

chú m lộn cak

mnjnnn 
  
  
5 tháng 10 2020

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

= ( x - 1 )( 2x + 1 )[ 1 + 3( x + 2 ) ]

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

= ( x - 1 )( 2x + 1 )( 3x + 7 )

b) ( 6x + 3 ) - ( 2x - 5 )( 2x + 1 )

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

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

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

= ( 2x + 1 )( 8 - 2x )

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

c) ( x - 5 )2 + ( x + 5 )( x - 5 ) - ( 5 - x )( 2x + 1 )

= ( x - 5 )2 + ( x + 5 )( x - 5 ) + ( x - 5 )( 2x + 1 )

= ( x - 5 )[ ( x - 5 ) + ( x + 5 ) + ( 2x + 1 ) ]

= ( x - 5 )( x - 5 + x + 5 + 2x + 1 )

= ( x - 5 )( 4x + 1 )

d) ( 3x - 2 )( 4x - 3 ) - ( 2 - 3x )( x - 1 ) - 2( 3x - 2 )( x + 1 )

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

= ( 3x - 2 )[ ( 4x - 3 ) + ( x - 1 ) - 2( x + 1 ) ]

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

= ( 3x - 2 )( 3x - 6 )

= 3( 3x - 2 )( x - 2 )

10 tháng 9 2021

jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

14 tháng 2 2020

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

\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)

\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)

\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)

\(\Leftrightarrow16x+52=0\)

\(\Leftrightarrow x=-\frac{13}{4}\)

b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow\)Xem lại đề !

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

\(\Leftrightarrow x^2-x-x^2-x+12=5x\)

\(\Leftrightarrow-2x+12=5x\)

\(\Leftrightarrow7x-12=0\)

\(\Leftrightarrow x=\frac{12}{7}\)

d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)

\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)

\(\Leftrightarrow28x+3x-1=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow x=\frac{1}{31}\)

14 tháng 2 2020

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

<=> 4x^2 + 12x + 9 - 3(x^2 - 16) = x^2 - 4x + 4 + 1 

<=> 4x^2 + 12x + 9 - 3x^2 + 48 = x^2 - 4x + 5

<=> x^2 + 12x + 57 = x^2 - 4x + 5

<=> x^2 - x^2 + 12x + 4x + 57 - 5 = 0

<=> 16x + 52 = 0

<=> 16x = -52

<=> x = -13/4

14 tháng 10 2018

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

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

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

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

2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)

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

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

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

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

\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)

\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)

\(\left(4-x\right)\left(5x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)

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

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

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

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

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

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

5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)

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

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

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

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

6) \(2x^2-5x-7=0\)

\(2x^2+2x-7x-7=0\)

\(2x\left(x+1\right)-7\left(x+1\right)=0\)

\(\left(x+1\right)\left(2x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)

7) \(x^2-x-12=0\)

\(x^2+3x-4x-12=0\)

\(x\left(x+3\right)-4\left(x+3\right)\)

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

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)

8) \(3x^2+14x-5=0\)

\(3x^2+15x-x-5=0\)

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

\(\left(x+5\right)\left(3x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)