A.0

thay số x= -2 và y= 1 ta có:

\(-2\left(-2\right)-4.1=4-4=0\)

\(\Rightarrow-2x-4y=0\)

\(\Rightarrow A\)

a)/x-2/+/x-5/=3
TH1:   

x-2+x-5=3
x+x-2-5=3
     2x-7=3
        2x=3+7
        2x=10
          x=10:2
          x=5
TH2

x-2+x-5= -3
x+x-2-5=-3
     2x-7=-3
        2x=-3+7
        2x=4
          x=4:2
          x=2
Vậy x\(\in\){5;2}

`|2x - 6| = 2x - 6 ∀ x >=3`

`|2x - 6| = 6 - 2x ∀ x < 3`.

`TH_1: x >=3`.

`<=> 2x - 6 = x + 4`.

<=> 2x - x = 4 + 6`.

`<=> x = 10 ( tm)`.

`TH_2: x < 3`.

`<=> 6 - 2x = x + 4`.

`<=> 2 = 3x`.

`<=> x = 2/3 ( tm )`.

Vậy tập nghiệm của phương trình là: `2/3, 10`.

b,|1-2x|+|5-2x|=4

==>|1-2x|+|5-2x|=|1-2x|+|2x-5|

==>|1-2x|+|2x-5| \(\geq\) |1-2x+2x-5|=|-4|=4

A \(\geq\) 4 khi \(\hept{\begin{cases}1-2x\\2x-5\end{cases}}\)\(\geq\) 0 ==>1/2\(\leq \) x \(\leq \) 5/2

giúp mình vs ạ

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

\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)

\(\Leftrightarrow-32x-26=0\)

\(\Leftrightarrow-32x=26\)

\(\Rightarrow x=-\frac{13}{16}\)

b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)

\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)

\(\Leftrightarrow16x^2+4x-3=0\)

\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)

c) \(\left(x+5\right)^2=45+x^2\)

\(\Leftrightarrow x^2+10x+25-x^2-45=0\)

\(\Leftrightarrow10x-20=0\)

\(\Leftrightarrow10x=20\)

\(\Rightarrow x=2\)

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

\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)

\(\Leftrightarrow-8x+11=0\)

\(\Leftrightarrow-8x=-11\)

\(\Rightarrow x=\frac{11}{8}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)

a) ( 2x - 3 )² - ( 2x + 5 )² = 10

<=> 4x² - 12x + 9 - ( 4x² + 20x + 25 ) = 10 

<=> 4x² - 12x + 9 - 4x² - 20x - 25 = 10

<=> -32x - 16 = 10

<=> -32x = 26

<=> x = -26/32 = -13/16

b) 4( x + 1 )² + ( 2x - 1 )² + 8( x + 1 )( x - 1 ) = 11

<=> 4( x² + 2x + 1 ) + ( 4x² - 4x + 1 ) + 8( x² - 1 ) = 11

<=> 4x² + 8x + 4 + 4x² - 4x + 1 + 8x² - 8 = 11

<=> 16x² + 4x - 14 = 0

<=> 2( 8x² + 2x - 7 ) = 0

<=> 8x² + 2x - 7 = 0

\(\Delta=b^2-4ac=2^2-4\cdot8\cdot\left(-7\right)=4+224=228\)( không muốn xài delta đâu nhưng bí quá ;-; )

\(\Delta>0\)nên phương trình đã cho có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-2+\sqrt{228}}{2\cdot8}=\frac{-2+\sqrt{228}}{16}=\frac{-1+\sqrt{57}}{8}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-2-\sqrt{228}}{2\cdot8}=\frac{-2-\sqrt{228}}{18}=\frac{-1-\sqrt{57}}{8}\end{cases}}\)

( Mình nghĩ ý này bạn nên xem lại đề )

c) ( x + 5 )² = 45 + x²

<=> x² + 10x + 25 = 45 + x²

<=> x² + 10x - x² = 45 - 25

<=> 10x = 20

<=> x = 2 

d) ( 2x - 3 )² - ( 2x - 1 )² = -3

<=> 4x² - 12x + 9 - ( 4x² - 4x + 1 ) = -3

<=> 4x² - 12x + 9 - 4x² + 4x - 1 = -3

<=> -8x + 8 = -3

<=> -8x = -11

<=> x = 11/8

e) ( x - 1 )² - ( 5x - 3 )² = 0

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

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

<=> [ -4x + 2 ][ 6x - 4 ] = 0

<=> \(\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)

(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0

=> 2x + 1 + 2x + 2 + ... + 2x + 2015 = 0

=> 2015 . 2x + (1 + 2 + ... + 2015) = 0

=> 4030x + (2015 + 1).2015 : 2 = 0

=> 4030x + 2031120 = 0

=> 4030x = -2031120

=> x = -504