Question

\(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)

Answer 1

\(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)

\(\Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\)

\(\Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\)

\(\Leftrightarrow64x=123\Leftrightarrow x=\dfrac{123}{64}\)

Answer 2

\(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)

=> \(\dfrac{3\left(4x^2-9\right)}{8}=\dfrac{4\left(x^2-8x+16\right)}{24}+\dfrac{8\left(x^2-4x+4\right)}{24}\)

=> 12x² - 27 = 4x² - 32x + 64 + 8x² - 32x + 32

=> 12x² - 27 = 12x² - 64x + 96

=> 64x = 123

=> x = \(\dfrac{123}{64}\)

Vậy,....

Answer 3

Đề là j thế bn ??? Mk làm giải Phương Trình nhé

\(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)

=> \(\dfrac{3\left(4x^2-9\right)}{24}=\dfrac{4\left(x-4\right)^2}{24}+\dfrac{8\left(x-2\right)^2}{24}\)

=> 12x² - 27 = 4( x² - 8x + 16 ) + 8( x² - 4x + 4)

=> 12x² - 17 = 4x² - 32x - 64 + 8x² - 32x + 32

=> 12x² - 17 = 12x² - 64x - 32

=> -64x - 32 = -17

=> -64x = 15

=> x = \(\dfrac{-15}{64}\)

Vậy , tập nhiệm của phương trình S = { \(\dfrac{-15}{64}\)}