tham khảo:
x+y+z+8=2√x−1+4√y−2+6√z−3a)x+y+z+8=2x-1+4y-2+6z-3 ĐK: x≥1;y≥2;z≥3x≥1;y≥2;z≥3
⇔x+y+z+8−2√x−1−4√y−2−6√z−3=0⇔x+y+z+8-2x-1-4y-2-6z-3=0
⇔(x−1−2√x−1+1)+(y−2−4√y−2+4)+(z−3−6√z−3+9)=0⇔(x-1-2x-1+1)+(y-2-4y-2+4)+(z-3-6z-3+9)=0
⇔(√x−1−1)2+(√y−2−2)2+(√z−3−3)2=0⇔(x-1-1)2+(y-2-2)2+(z-3-3)2=0
Do (√x−1−1)2≥0;(√y−2−2)2≥0;(√z−3−3)2≥0(x-1-1)2≥0;(y-2-2)2≥0;(z-3-3)2≥0
⇒(√x−1−1)2+(√y−2−2)2+(√z−3−3)2≥0⇒(x-1-1)2+(y-2-2)2+(z-3-3)2≥0
Dấu = xảy ra khi ⎧⎪ ⎪⎨⎪ ⎪⎩√x−1=1√y−2=2√z−3=3⇔⎧⎪⎨⎪⎩x−1=1y−2=4z−3=9⇔⎧⎪⎨⎪⎩x=2(tm)y=6(tm)z=12(tm){x-1=1y-2=2z-3=3⇔{x-1=1y-2=4z-3=9⇔{x=2(tm)y=6(tm)z=12(tm)
Vậy (x;y;z)=(2;6;12)
ĐK: \(x\ge1;y\ge2;z\ge3\)
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=6\left(tm\right)\\z=12\left(tm\right)\end{matrix}\right.\)
