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ĐKXĐ : \(\hept{\begin{cases}x\ge1\\y\ge2\\z\ge3\end{cases}}\)
Với điều kiện trên thì pt đã cho tương đương với :
\(\left[\left(x-1\right)-2\sqrt{x-1}+1\right]+\left[\left(y-2\right)-4\sqrt{y-2}+4\right]+\left[\left(z-3\right)-6\sqrt{z-3}+9\right]=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
Mà \(\left(\sqrt{x-1}-1\right)^2\ge0,\left(\sqrt{y-2}-2\right)^2\ge0,\left(\sqrt{z-3}-3\right)^2\ge0\)
\(\Rightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2\ge0\)
Vậy đẳng thức xảy ra khi \(\hept{\begin{cases}\left(\sqrt{x-1}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{z-3}-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\) (tmđk)
ĐKXĐ : {
| x≥1 |
| y≥2 |
| z≥3 |
Với điều kiện trên thì pt đã cho tương đương với :
[(x−1)−2√x−1+1]+[(y−2)−4√y−2+4]+[(z−3)−6√z−3+9]=0
⇔(√x−1−1)2+(√y−2−2)2+(√z−3−3)2=0
Mà (√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
Vậy đẳng thức xảy ra khi {
| (√x−1−1)2=0 |
| (√y−2−2)2=0 |
| (√z−3−3)2=0 |
Sai đề kìa \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x-1}+1-1\right)+\left(y-4\sqrt{y-2}+4-2\right)+\left(z-6\sqrt{z-3}+9-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
Sai đề kìa x+y+z+8=2√x−1+4√y−2+6√z−3
⇔x+y+z+8−2√x−1−4√y−2−6√z−3=0
⇔(x−2√x−1+1−1)+(y−4√y−2+4−2)+(z−6√z−3+9−3)=0
⇔(√x−1−1)2+(√y−2−2)2+(√z−3−3)2=0
⇒{
| √x−1−1=0 |
| √y−2−2=0 |
| √z−3−3=0 |
⇒{
| √x−1=1 |
| √y−2=2 |
| √z−3=3 |
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Rightarrow\left(x-1\right)-2\sqrt{x-1}+1\)\(+\left(y-2\right)-4\sqrt{y-2}+4\)\(+\left(z-3\right)-6\sqrt{z-3}+9\)\(=0\)
\(\Rightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}}\)
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-2\sqrt{y-2}.2+4\right)+\left(z-3-2\sqrt{z-3}.3+9\right)=0\)
\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)( 1 )
Mà \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2\ge0\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\left(\sqrt{x-1}-1\right)^2=\left(\sqrt{y-2}-2\right)^2=\left(\sqrt{z-3}-3\right)^2=0\)
từ đó tìm được : \(x=2;y=6;z=12\)
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow\left[\left(x-1\right)-2\sqrt{x-1}+1\right]+\left[\left(y-2\right)-4\sqrt{y-2}+4\right]+\left[\left(z-3\right)-6\sqrt{z-3}+9\right]=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
ĐK: \(x\ge1,y\ge2,z\ge3\).
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)(thỏa mãn)
\(A=\sqrt{x^3+8}+\sqrt{y^3+8}+\sqrt{z^3+8}\)
\(A=\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2x+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}\)
\(\sqrt{\frac{1}{2}}A=\sqrt{\left(x+2\right)\left(x^2-2x+4\right).\frac{1}{2}}+\sqrt{\left(y+2\right)\left(y^2-2x+4\right).\frac{1}{2}}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right).\frac{1}{2}}\)\(\sqrt{\frac{1}{2}}A=\sqrt{\left(x+2\right)\left(\frac{x^2}{2}-x+2\right)}+\sqrt{\left(y+2\right)\left(\frac{y^2}{2}-x+2\right)}+\sqrt{\left(z+2\right)\left(\frac{z^2}{2}-z+2\right)}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{\frac{1}{2}}A\le\frac{x+2+\frac{x^2}{2}-x+2+y+2+\frac{y^2}{2}-y+2+z+2+\frac{z^2}{2}-z+2}{2}=\frac{12+\frac{x^2+y^2+z^2}{2}}{2}=\frac{12+\frac{48}{2}}{2}=\frac{12+24}{2}=\frac{36}{2}=18\)
\(\Leftrightarrow A\le18:\sqrt{\frac{1}{2}}=18\sqrt{2}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2=\frac{x^2}{2}-x+2\\y+2=\frac{y^2}{2}-y+2\\z+2=\frac{z^2}{2}-z+2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=4x\\y^2=4y\\z^2=4z\end{cases}}\Leftrightarrow\hept{\begin{cases}x\left(x-4\right)=0\\y\left(y-4\right)=0\\z\left(z-4\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=4\\z=4\end{cases}\left(v\text{ì}x,y,z>0\right)}}\)
Vậy \(A_{max}=18\sqrt{2}\Leftrightarrow x=y=z=4\)
Tham khảo nhé~
Giải xong rồi
(x;y;z)=(2;6;12)
OK
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