Tìm x 1, (x-1)(x^2 + x +1) - x(x+2)(x-2) = 5 2, (x+1)^2 - 2(x+1)(3x -2) + (3x -2)^2 =0 3, x^2 - 2x +1=0 4, (5x +1)^2 - (5x -3)(5x+3)=30

Đáp án:

Giải thích các bước giải:

`\text{@Terminators}`

`1, (x-1)(x^2+x+1)-x(x+2)(x-2)=5`

`-> x^3 - 1 - x^3 + 4x = 5`

`-> 4x - 1 = 5`

`-> 4x = 6`

`-> x = 3/2`

Vậy `x \in {3/2}`

`2, (x+1)^2 - 2(x+1)(3x-2)+(3x-2)^2=0` 

`-> x^2 + 2x + 1 - 6x^2 - 2x + 4 + 9x^2 - 12x + 4 = 0`

`-> 4x^2 - 12x + 9 = 0`

`-> (2x)^2 - 2 * 2x * 3 + 3^2 = 0`

`-> (2x-3)^2 = 0`

`-> (2x-3)^2 = 0^2`

`-> 2x - 3 = 0`

`-> 2x =3`

`-> x = 3/2`

Vậy `x \in {3/2}`

`3, x^2 - 2x + 1 = 0`

`-> x^2 - 2x * 1 + 1^2 = 0`

`-> (x-1)^2 = 0`

`-> (x-1)^2 = 0^2`

`-> x - 1 = 0`

`-> x = 1`

Vậy `x \in {1}`

`4, (5x+1)^2-(5x-3)(5x+3)=30`

`-> 25x^2 + 10x + 1 - 25x^2 + 9 = 30`

`-> 10x + 10 = 30`

`-> 10x = 20`

`-> x = 2`

Vậy `x \in {2}`

Đáp án:

Giải thích các bước giải: \(\begin{array}{l} 1.\,\,\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - x\left( {x + 2} \right)\left( {x - 2} \right) = 5\\ \Leftrightarrow {x^3} - 1 - x\left( {{x^2} - 4} \right) = 5\\ \Leftrightarrow 4x - 1 = 5\\ \Leftrightarrow 4x = 6\\ \Leftrightarrow x = \frac{3}{2}\\ 2.\,\,\,{\left( {x + 1} \right)^2} - 2\left( {x + 1} \right)\left( {3x - 2} \right) + {\left( {3x - 2} \right)^2} = 0\\ \Leftrightarrow {\left( {x + 1 - 3x + 2} \right)^2} = 0\\ \Leftrightarrow - 2x + 3 = 0\\ \Leftrightarrow x = \frac{3}{2}\\ 3.\,\,\,{x^2} - 2x + 1 = 0\\ \Leftrightarrow {\left( {x - 1} \right)^2} = 0\\ \Leftrightarrow x - 1 = 0\\ \Leftrightarrow x = 1\\ 4.\,\,{\left( {5x + 1} \right)^2} - \left( {5x - 3} \right)\left( {5x + 3} \right) = 30\\ \Leftrightarrow 25{x^2} + 10x + 1 - 25{x^2} + 9 = 30\\ \Leftrightarrow 10x + 10 = 30\\ \Leftrightarrow 10x = 20\\ \Leftrightarrow x = 2 \end{array}\)