Question

a)(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0

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

c)(x+3)^3-x(3x+1)^2+(2x-1)(4x^2-2x+1)=28

d)(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0

Answer 1

_{Sorry mình nhầm câu a}

a) (2x - 1)² + (x + 3)² - 5(x + 7)(x - 7) = 0

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

c) (x + 3)³ - x(3x + 1)² + (2x - 1)(4x² - 2x + 1) = 28

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

Giải:

a) (2x - 1)² + (x + 3)² - 5(x + 7)(x - 7) = 0

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5(x² - 49) = 0

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

\(\Leftrightarrow\) 2x + 255 = 0

\(\Leftrightarrow\) 2x = - 255

\(\Leftrightarrow\) x = - 255 : 2

\(\Leftrightarrow\) x = \(-\frac{255}{2}\)

Vậy x = \(-\frac{255}{2}\)

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

\(\Leftrightarrow\) x³ + 8 - x³ - 2x = 15

\(\Leftrightarrow\) 8 - 2x = 15

\(\Leftrightarrow\) 2x = 8 - 1

\(\Leftrightarrow\) 2x = - 7

\(\Leftrightarrow\) x = - 7 : 2

\(\Leftrightarrow\) x = \(-\frac{7}{2}\)

Vậy x = \(-\frac{7}{2}\)

c) (x + 3)³ - x(3x + 1)² + (2x - 1)(4x² - 2x + 1) = 28

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - x(9x² + 6x + 1) + 8x³ - 1 = 28

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - 9x³ - 6x² - x + 8x³ - 1 = 28

\(\Leftrightarrow\) 26x + 26 = 28

\(\Leftrightarrow\) 26x = 28 - 26

\(\Leftrightarrow\) 26x = 2

\(\Leftrightarrow\) x = 2 : 26

\(\Leftrightarrow\) x = \(\frac{1}{13}\)

Vậy x = \(\frac{1}{13}\)

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - (x⁶ - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - x⁶ + 1 = 0

\(\Leftrightarrow\) -2x² + 2 = 0

\(\Leftrightarrow\) -2x² = - 2

\(\Leftrightarrow\) x² = - 2 : (- 2)

\(\Leftrightarrow\) x² = 1

\(\Leftrightarrow\) x = 1 hoặc x = - 1

Vậy x \(\in\) {1; - 1}

Answer 2

cơ may sao nay gặp được Dân Chơi Đất Bắc luôn nè ::)))