+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)

\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)

+) \(\dfrac{4}{3}x=-\dfrac{2}{3}+\dfrac{1}{2}=-\dfrac{1}{6}\)

\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)

+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{31}{12}\)

\(\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{4}{3}\)

\(\dfrac{1}{3}x=\left(-\dfrac{4}{3}\right)-\dfrac{1}{2}\)

\(\dfrac{1}{3}x=-\dfrac{11}{6}\)

\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)

\(x=-\dfrac{11}{2}\)

\(-\dfrac{2}{3}-\dfrac{4}{3}x=-\dfrac{1}{2}\)

\(\dfrac{4}{3}x=\left(-\dfrac{2}{3}\right)-\dfrac{-1}{2}\)

\(\dfrac{4}{3}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)

\(x=-\dfrac{1}{8}\)

\(\dfrac{5}{2}-2\left(x-1\right)=-\dfrac{2}{3}\)

\(2\left(x-1\right)=\dfrac{5}{2}-\left(-\dfrac{2}{3}\right)\)

\(2\left(x-1\right)=\dfrac{19}{6}\)

\(\left(x-1\right)=\dfrac{19}{6}:2\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{19}{12}+1\)

\(x=\dfrac{31}{12}\)

\(\dfrac{7}{-2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x=\left(-\dfrac{1}{2}\right)+\dfrac{1}{3}\)

\(\dfrac{-7}{2}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\left(-\dfrac{7}{2}\right)\)

\(x=\dfrac{1}{21}\)

\(\dfrac{8}{5}-\dfrac{1}{2}:x=\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=\dfrac{8}{5}-\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=-\dfrac{16}{15}\)

\(x=\dfrac{1}{2}:\left(-\dfrac{16}{15}\right)\)

\(x=-\dfrac{15}{32}\)

\(-\dfrac{5}{4}x-\dfrac{1}{2}x=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{5}{4}-\dfrac{1}{2}\right)=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{7}{4}\right)=-\dfrac{7}{3}\)

\(x=\left(-\dfrac{7}{3}\right):\left(-\dfrac{7}{4}\right)\)

\(x=\dfrac{4}{3}\)

a: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1=28\)

\(\Leftrightarrow9x^3+9x^2+27x+28-28-9x^3-6x^2-x=0\)

\(\Leftrightarrow3x^2+26x=0\)

=>x=0 hoặc x=-26/3

b: \(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)

\(\Leftrightarrow-3x^4+3x^2=0\)

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

hay \(x\in\left\{0;1;-1\right\}\)

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

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

=> \(x^6-3x^4+3x^2-1-x^6+1=0\)

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

=> \(-3x^4+3x^2=0\)

=> \(-\left(3x^4-3x^2\right)=0\)

=> \(3x\left(x^3-x\right)=0\)

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

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

_{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 ::)))

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...