Bài 1: Tìm x, biết: a) (x+1) mũ 2 -1 =0 b) x mũ 3 +3x -1 =0 c) x mũ 2 + 4x=21 d) (2x-1) mũ 2 + (x+3) mũ 2 -5(x+7)(x-7)=0 e) x mũ 3+ x mũ 2+ 1 phần 3x+

Đáp án:

1. 

a, `(x + 1)^2 – 1 = 0`

`<=> (x + 1)^2 = 1`

<=>  \(\left[ \begin{array}{l}x + 1 = 1\\x + 1 = -1\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\) 

b, `x^3 + 3x – 1 = 0` < Dề sai>

c, `x^2 + 4x = 21`

`<=> x^2 + 4x + 4 = 25`

`<=> (x + 2)^2  =25`

<=> \(\left[ \begin{array}{l}x + 2 = 5\\x + 2 = -5\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x=3\\x=-7\end{array} \right.\) 

d, `(2x – 1)^2 + (x + 3)^2 – 5(x + 7)(x – 7) = 0`

`<=> 4x^2 – 4x + 1 + x^2 + 6x + 9 – 5x^2 + 245 = 0`

`<=> 2x + 255 = 0`

`<=> 2x = -255`

`<=> x = -255/2`

e, `x^3 + x^2 + 1/3 x + 1/27 = 0`

`<=> x^3 + 3.x^2 . 1/3 + 3.x . (1/3)^2 + (1/3)^3 = 0`

`<=> (x + 1/3)^3 = 0`

`<=> x + 1/3 = 0`

`<=> x = -1/3`

2. 

a, Ta có : 

c1 : Tính đỡ nhóc nhưng làm dài

`(x – y)^2 + (y – z)^2 + ( z – x)^2 = (y + z – 2x)^2 + (z + x – 2y)^2 + (x + y – 2z)^2`

`<=> [(x + y – 2z)^2 – (x – y)^2] + [(z + x – 2y)^2 – (z – x)^2] + [(y + z – 2x)^2 – (y – z)^2] = 0`

`<=> (x + y – 2z – x + y)(x + y – 2z + x – y) + (z + x – 2y – z + x)(z + x – 2y + z – x) + (y + z – 2x – y + z)(y + z – 2x + y – z) = 0`

`<=> (2y – 2z)(2x – 2z) + (2x – 2y)(2z – 2y) + (2z – 2x)(2y – 2x) = 0`

`<=> 4.(y – z)(x – z) + 4.(x – y)(z – y) + 4.(z – x)(y – x) = 0`

`<=> (y – z)(x – z) + (x – y)(z – y) + (z – x)(y – x) =  0`

`<=> [(y – z)(x – z) + (x – y)(z – y)] + [(x – y)(z – y) + (z – x)(y – x)] + [(z – x)(y – x) + (y – z)(x – z)] = 0`

`<=> [(y – z)(x – z) – (x – y)(y – z)] + [(x – y)(y – z) – (x – y)(z – x)] + [(z – x)(y – x) – (z – x)(y – z)]= 0`

`<=> (y – z)(x – z – x + y) + (x – y)(z – y – z + x) + (z – x)(y – x – y + z) = 0`

`<=> (y – z)^2 + (x – y)^2 + (z – x)^2 = 0`

`<=> y – z = x – y = z – x = 0`

`<=> x = y = z`

c2 Ta có : 

`(x – y)^2 + (y – z)^2 + ( z – x)^2 = (y + z – 2x)^2 + (z + x – 2y)^2 + (x + y – 2z)^2`

`<=> (y + z – 2x)^2 + (z + x – 2y)^2 + (x + y – 2z)^2 – (x – y)^2 –  (y – z)^2 –  ( z – x)^2 = 0`

`<=> y^2 + z^2 + 4x^2 + 2yz – 4zx – 4xy + z^2 + x^2 + 4y^2 + 2zx – 4xy – 4yz + x^2 + y^2 + 4z^2 + 2xy – 4yz – 4zx – x^2 + 2xy – y^2 – y^2 + 2yz – z^2 – z^2 + 2zx – x^2 = 0`

`<=> 4x^2 + 4y^2 + 4z^2 – 4xy – 4yz – 4zx = 0`

`<=> 8x^2 + 8y^2 + 8z^2 – 8xy – 8yz – 8zx = 0`

`<=> (4x^2 – 8xy + 4y^2) + (4y^2 – 8yz + 4z^2) + (4z^2 – 8zx + 4x^2) = 0`

`<=> (2x – 2y)^2 + (2y – 2z)^2 + (2z – 2x)^2 = 0`

`<=> 2x – 2y = 2y – 2z = 2z – 2x = 0`

`<=> x – y = y – z = z – x = 0`

`<=> x = y = z`

b, Ta có :

`(x + y)(x^2 – xy + y^2) + (x –  y)(x^2 + xy + y^2) – 2x^3`

`= x^3 + y^3 + x^3 – y^3 – 2x^3`

`= 2x^3 – 2x^3`

`= 0`

=> Biểu thức có giá trị không phụ thuộc vào biến

Giải thích các bước giải:

 

Trả lời