Question

giải các phương trình sau

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

b) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

c) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)

Answer 1

a, \(\Leftrightarrow3x^2-3+5=3x^2+2x-3x-2\)

\(\Leftrightarrow3x^2-3x-2x+3x=-2+3-5\)

<=>x=-4

b, \(\Leftrightarrow\dfrac{x+4}{5}-\dfrac{5x}{5}+\dfrac{20}{5}=\dfrac{2x}{6}-\dfrac{3\left(x-2\right)}{6}\)

\(\Leftrightarrow\dfrac{x+4-5x+20}{5}=\dfrac{2x-3x+6}{6}\)

\(\Leftrightarrow\dfrac{6\left(-4x+24\right)}{30}=\dfrac{5\left(-x+6\right)}{30}\)

<=>-24x+144=-5x+30

<=>-5x+24x=144-30

<=>19x=114

<=>x=6

Answer 2

a ) <=> 3x² - 3 + 5 = 3x² + 2x - 3x - 2

<=> 3x² - 3x² - 2x + 3x = -2 - 5 + 3

<=> x = - 4

Vậy s = \(\left\{-4\right\}\)

b)<=> \(\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> 6x -30x - 10x + 15x = 30 - 24 + 120

<=> -19x = 126

<=> x =-6,6

Vậy s = \(\left\{-6,6\right\}\)

Answer 3

a ) <=> 3x² - 3 + 5 = 3x² + 2x - 3x - 2

<=> 3x² - 3x² - 2x + 3x = -2 - 5 + 3

<=> x = - 4

Vậy s = \(\left\{-4\right\}\)

b)<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> -19x = 126

<=> x =-6,6

Answer 4

c, \(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{29-x+21}{21}+\dfrac{27-x+23}{23}+\dfrac{25-x+25}{25}+\dfrac{23-x+27}{27}+\dfrac{21-x+29}{29}=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

Mà 1/21+1/23+1/25+1/27+1/29 khác 0

<=>50-x=0<=>x=50