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+ 1 phần 27=0
Bài 2:
a) Chứng minh rằng nếu:
(x-y) mũ 2 + (y-z) mũ 2 + (z-x) mũ 2 = (y+z-2x) mũ 2 + (z+x-2y) mũ 2 + (x+y-2z) mũ 2 thì x=y=z
b) Chứng minh rằng giá trị của biểu thức sau không phụ thuộc vào x,y:
(x+y)(x mũ 2 -xy +y mũ 2) + (x+y)(x mũ 2 +xy+y mũ 2) – 2x mũ 3
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
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