Question

Tìm x:

a,(𝑥+2)2+(𝑥+3)2−2(𝑥−2)(𝑥−3)=19

b,(x+2)(x^2-2x+4)-x(x^2-5)=15

c, (x-1)^3+(2-x)(4+2x+x^2)+3x(x+2)=17

Answer 1

\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)

Answer 2

a. (x + 2)² + (x + 3)² - 2(x - 2)(x - 3) = 19

<=> (x² + 4x + 4) + (x² + 6x + 9) - (2x + 4)(x - 3) = 19

<=> x² + 4x + 4 + x² + 6x + 9 - 2x² + 6x - 4x + 12 = 19

<=> x² + x² - 2x² + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0

<=> 12x + 6 = 0

<=> 6(2x + 1) = 0

<=> 2x + 1 = 0

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)  

Answer 3

b. (x + 2)(x² - 2x + 4) - x(x² - 5) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 5x - 15 = 0

<=> x³ - x³ - 2x² + 2x² + 4x - 4x + 5x + 8 - 15 = 0

<=> 5x - 7 = 0

<=> 5x = 7

<=> x = \(\dfrac{7}{5}\)