Cho x+y+z=0 CMR (x^2+ y^2 + z^2)^2 = 2(x^4+ y^4 + z^4) 31/08/2021 Bởi Adalynn Cho x+y+z=0 CMR (x^2+ y^2 + z^2)^2 = 2(x^4+ y^4 + z^4)
Ta có: $x + y+ z = 0$ ⇒ $x = -(y+z)$ ⇒ $x^2 = [-(y+z)]^2$ ⇒ $x^2 = (y+z)^2$ ⇒ $x^2 = y^2 + 2yz + z^2$ ⇒ $x^2 – y^2 – z^2 = y^2 + 2yz + z^2 – y^2 – z^2 = 2yz$ Do đó: $(x^2 – y^2 – z^2)^2 = (2yz)^2 = 4y^2z^2$ ⇔ $x^4 + y^4 + z^4 – 2x^2y^2 – 2x^2z^2 + 2y^2z^2 = 4y^2z^2$ ⇔ $x^4 + y^4 + z^4 = 4y^2z^2 + 2x^2y^2 + 2x^2z^2 – 2y^2z^2$ ⇔ $x^4 + y^4 + z^4 = 2y^2z^2 + 2x^2y^2 + 2x^2z^2$ ⇔ $x^4 + y^4 + z^4 + x^4 + y^4 + z^4 = x^4 + y^4 + z^4 + 2[(yz)^2 + (xy)^2 + (xz)^2]$ ⇔ $2(x^4 + y^4 + z^4) = (x^2 + y^2 + z^2)^2$ Bình luận
Ta có: `x + y + z = 0` `⇒ x = -(y + x) ` `⇒ x² = (y + x)² ` `⇒ x² = y² + 2yz + z² ` `⇒ x² – y² – z² = 2yz ` `⇒ (x² – y² – z²)² = 4y²z²` ⇒ `x^4` + `y^4` + `z^4` `- 2x²y² – 2x²z² + 2y²z² = 4y²z² ` ⇒ `x^4` + `y^4` + `z^4` `= 4y²z² – 2y²z² + 2x²z² + 2x²y² ` `= 2x²y² + 2y²z² + 2x²z²` ⇒ 2(`x^4` + `y^4` + `z^4`) `= (x² + y² + z²)²` Bình luận
Ta có: $x + y+ z = 0$
⇒ $x = -(y+z)$
⇒ $x^2 = [-(y+z)]^2$
⇒ $x^2 = (y+z)^2$
⇒ $x^2 = y^2 + 2yz + z^2$
⇒ $x^2 – y^2 – z^2 = y^2 + 2yz + z^2 – y^2 – z^2 = 2yz$
Do đó: $(x^2 – y^2 – z^2)^2 = (2yz)^2 = 4y^2z^2$
⇔ $x^4 + y^4 + z^4 – 2x^2y^2 – 2x^2z^2 + 2y^2z^2 = 4y^2z^2$
⇔ $x^4 + y^4 + z^4 = 4y^2z^2 + 2x^2y^2 + 2x^2z^2 – 2y^2z^2$
⇔ $x^4 + y^4 + z^4 = 2y^2z^2 + 2x^2y^2 + 2x^2z^2$
⇔ $x^4 + y^4 + z^4 + x^4 + y^4 + z^4 = x^4 + y^4 + z^4 + 2[(yz)^2 + (xy)^2 + (xz)^2]$
⇔ $2(x^4 + y^4 + z^4) = (x^2 + y^2 + z^2)^2$
Ta có:
`x + y + z = 0`
`⇒ x = -(y + x) `
`⇒ x² = (y + x)² `
`⇒ x² = y² + 2yz + z² `
`⇒ x² – y² – z² = 2yz `
`⇒ (x² – y² – z²)² = 4y²z²`
⇒ `x^4` + `y^4` + `z^4` `- 2x²y² – 2x²z² + 2y²z² = 4y²z² `
⇒ `x^4` + `y^4` + `z^4` `= 4y²z² – 2y²z² + 2x²z² + 2x²y² `
`= 2x²y² + 2y²z² + 2x²z²`
⇒ 2(`x^4` + `y^4` + `z^4`) `= (x² + y² + z²)²`