Question

Giải Phương trình

a) (5,5-11x)(\(\frac{7x+2}{5}\)+ (\(\frac{2\left(1-3x\right)}{3}\))= 0

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

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

d) (3x^2+1)(4x-3)= (3x^2+1)(x-7)

Answer 1

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

<=> 6x² - 11x + 3 = 2x² - x - 6

<=.> 4x² - 10x + 9 = 0

<=> (2x - \(\frac{5}{2}\))² +\(\frac{11}{4}\)= 0 ( vô lí )

( Vì (2x - \(\frac{5}{2}\))² \(\ge\) 0 => (2x - \(\frac{5}{2}\))² + \(\frac{11}{4}\)\(\ge\)\(\frac{11}{4}\))

Vậy pt vô nghiệm

Answer 2

câu C : (4-3x)(2x+3)=(5-2x)(3x-4)

<=> (4-3x)(2x+3)-(5-2x)(4-3x)=0

<=>(4-3x)(2x+3-5+2x)=0

<=>(4-3x)(4x-2)=0

<=>\(\left[\begin{matrix}3x=4\\4x=2\end{matrix}\right.\)

<=>\(\left[\begin{matrix}x=\frac{4}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Answer 3

d,

(3x² +1)(4x - 3 ) = ( 3x² +1 )(x - 7)

<=> (3x² + 1 )( 4x - 3) - ( 3x² + 1 )( x - 7 ) =0

<=> (3x² +1 )\(\left[\left(4x-3\right)-\left(x-7\right)\right]\)= 0

<=> (3x² + 1 ) ( 3x + 4 ) = 0

<=> 3x + 4 = 0

<=> x = -\(\frac{4}{3}\)

Vậy pt có nghiệm là - \(\frac{4}{3}\)

Answer 4

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

<=> 8x + 12 - 6x² - 9x = 15x - 20 - 6x² + 8x

<=> 12 - 9x - 15x + 20 = 0

<=> 32 - 24x = 0

<=> 8(4 - 3x) = 0

<=> 4 - 3x = 0

<=> x = \(\frac{4}{3}\)

Vậy pt có nghiệm là x = \(\frac{4}{3}\)

Answer 5

bai nay ha