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

=> x³ + 9x² + 27x + 27 - x(9x² + 6x + 1) +(2x + 1)[(2x)² - 2.x.1 + 1² ] = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + (2x)³ + 1³ = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

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

=> 3x² + 26x + 28 = 28

=> 3x² + 26x = 0

=> 3x² + 26x = 0

=> \(3x\left(x+\frac{26}{3}\right)=0\)

=> 3x = 0 hoặc x + 26/3 = 0

=> x = 0 hoặc x = -26/3

b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-x^6+1=0\)

=> \(\left(x^6-x^6\right)-3x^4+3x^2+\left(-1+1\right)=0\)

=> \(-3x^4+3x^2=0\)

=> \(-\left(3x^4-3x^2\right)=0\)

=> \(3x\left(x^3-x\right)=0\)

=> \(\orbr{\begin{cases}3x=0\\x^3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x\left(x^2-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=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}

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

a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)

\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)

\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

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