Giúp mk với ạ !! x ³ +y ³+z ³=3xyz .Chứng minh x=y=z 11/11/2021 Bởi Nevaeh Giúp mk với ạ !! x ³ +y ³+z ³=3xyz .Chứng minh x=y=z
$\quad x^3 + y^3 + z^3 = 3xyz$ $\to x^3 + y^3 + z^3 – 3xyz=0$ $\to (x+y)^3 – 3xy(x+y)+ z^3 – 3xyz=0$ $\to (x+y+z)[(x+y)^2 – z(x+y) + z^2] – 3xy(x+y+z)=0$ $\to (x+y+z)(x^2 + 2xy + y^2 – zx – yz + z^2 – 3xy)=0$ $\to (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx)=0$ $\to (x+y+z)(2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx)=0$ $\to (x+y+z)[(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 – 2zx + x^2)]=0$ $\to (x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]=0$ $\to \left[\begin{array}{l}x + y + z = 0\\x = y = z\end{array}\right.$ Bình luận
Đáp án: Giải thích các bước giải: x³+y³+z³=3xyz ⇒ x³+y³+z³-3xyz=0 ⇔ (x+y)³-3x²y-3xy²+z³-3xyz=0 ⇔ [(x+y)³+z³]-3xy(x+y+z)=0 ⇔ (x+y+z)[(x+y)²-z(x+y)+z²]-3xy(x+y+z)=0 ⇔ (x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)=0 ⇔ (x+y+z)(x²+y²+z²-xy-yz-xyz)=0 ⇒ x+y+z=0 hoặc x²+y²+z²-xy-yz-xz=0 ta có x²+y²+z²-xy-yz-xz=0 ⇒ 2x²+2y²+2z²-2xy-2yz-2xz=0 ⇔ (x²+y²-2xy)+(y²+z²-2yz)+(z²+x²-2xz)=0 ⇔ (x-y)²+(y-z)²(z-x)²=0 ⇒ x=x=z Bình luận
$\quad x^3 + y^3 + z^3 = 3xyz$
$\to x^3 + y^3 + z^3 – 3xyz=0$
$\to (x+y)^3 – 3xy(x+y)+ z^3 – 3xyz=0$
$\to (x+y+z)[(x+y)^2 – z(x+y) + z^2] – 3xy(x+y+z)=0$
$\to (x+y+z)(x^2 + 2xy + y^2 – zx – yz + z^2 – 3xy)=0$
$\to (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx)=0$
$\to (x+y+z)(2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx)=0$
$\to (x+y+z)[(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 – 2zx + x^2)]=0$
$\to (x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]=0$
$\to \left[\begin{array}{l}x + y + z = 0\\x = y = z\end{array}\right.$
Đáp án:
Giải thích các bước giải:
x³+y³+z³=3xyz
⇒ x³+y³+z³-3xyz=0
⇔ (x+y)³-3x²y-3xy²+z³-3xyz=0
⇔ [(x+y)³+z³]-3xy(x+y+z)=0
⇔ (x+y+z)[(x+y)²-z(x+y)+z²]-3xy(x+y+z)=0
⇔ (x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)=0
⇔ (x+y+z)(x²+y²+z²-xy-yz-xyz)=0
⇒ x+y+z=0 hoặc x²+y²+z²-xy-yz-xz=0
ta có x²+y²+z²-xy-yz-xz=0
⇒ 2x²+2y²+2z²-2xy-2yz-2xz=0
⇔ (x²+y²-2xy)+(y²+z²-2yz)+(z²+x²-2xz)=0
⇔ (x-y)²+(y-z)²(z-x)²=0
⇒ x=x=z