Question

Tìm x:
a. \(\frac{4}{x+3}=\frac{3}{2x-1}\)
b.\(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)
c. \(\left(2x-1\right)^2=\left(2x-1\right)^3\)

Answer 1

a) \(\frac{4}{x+5}=\frac{3}{2x-1}\)

=> 4(2x - 1) = 3(x + 5)

=> 8x - 4 = 3x + 15

=> 8x - 3x = 15 + 4

=> 5x = 19

=> x = 19/5

b) \(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)

=> \(\left(\frac{x+11}{19}-1\right)+\left(\frac{x+12}{20}-1\right)+\left(\frac{x+13}{21}-1\right)=0\)

=> \(\frac{x-8}{19}+\frac{x-8}{20}+\frac{x-8}{21}=0\)

=> \(\left(x-8\right)\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)

=> x - 8 = 0

=> x = 8

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

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

=> \(\left(2x-1\right)^2.\left[1-\left(2x-1\right)\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)=0\end{cases}}\)

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

=> \(\orbr{\begin{cases}2x=1\\2-2x=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

Answer 2

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

<=> 4.(2x - 1) = (x + 3).3

<=> 8x - 4 = 3x + 9

<=> 8x = 3x + 9 + 4

<=> 8x = 3x + 13

<=> 8x - 3x = 13

<=> 5x = 13

<=> x = 13/5

=> x = 13/5

c) (2x - 1)² = (2x - 1)³

<=> 4x² - 4x + 1 = 8x³ - 12x² + 6x - 1

<=> 8x³ - 12x² + 6x - 1 = 4x² - 4x + 1

<=> 8x³ - 12x² + 6x - 1 - 1 = 4x² - 4x

<=> 8x³ - 12x² + 6x - 2x = 4x² - 4x

<=> 8x³ - 12x² + 6x - 2x - 4x = 4x²

<=> 8x³ - 12x² + 10x - 2 = 4x²

<=> 8x³ - 12x² + 10x - 2 - 4x² = 0

<=> 8x² - 16x² + 10x - 2 = 0

<=> 2(x - 1)(2x - 1)² = 0

<=> x - 1 = 0 hoặc 2x - 1 = 0

       x = 0 + 1         2x = 0 + 1

       x = 1               2x = 1

                              x = 1/2

=> x = 1 hoặc x = 1/2