Question

[Phi DU]

Giải giúp mình gấp nha. Cảm ơn mấy bạn.

a) x²+y²+z²=x(y+z)

b) (x+y)²=(x+1)(y-1)

c) x²+y²+z²+t²=x(y+z+t)

d) (x²+1)(y+4)(z²+9)=48xyz

e) (x+1)(y+1)(x+y)=8xy (x+y>=0)

Answer 1

[Lightning Farron]

d)Áp dụng BĐT AM-GM

\(x^2+1\ge2\sqrt{x^2}=2x\)

\(y^2+4\ge2\sqrt{4y^2}=4y\)

\(z^2+9\ge2\sqrt{9z^2}=6z\)

Nhân theo vế ta có:

\(VT=\left(x^2+1\right)\left(y^2+4\right)\left(z^2+9\right)\ge2x\cdot4y\cdot6z=48xyz=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x^2+1=2x\\y^2+4=4y\\z^2+9=6z\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

e)Áp dụng BĐT AM-GM ta có:

\(x+1\ge2\sqrt{x}\)

\(y+1\ge2\sqrt{y}\)

\(x+y\ge2\sqrt{xy}\)

Nhân theo vế ta có:

\(VT=\left(x+1\right)\left(y+1\right)\left(x+y\right)\ge2\sqrt{x}\cdot2\sqrt{x}\cdot2\sqrt{xy}=8xy=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x+1=2\sqrt{x}\\y+1=2\sqrt{y}\\x+y=2\sqrt{xy}\left(x+y\ge0\right)\end{matrix}\right.\)\(\Rightarrow x=y=0\)

Answer 2

[Lightning Farron]

mấy câu còn lại áp dụng HĐT thôi, khá dễ !!