\(c^2+d^2+25=6c+8d\)

\(\Leftrightarrow\left(c^2-6c+9\right)+\left(d^2-8d+16\right)=0\)

\(\Leftrightarrow\left(c-3\right)^2+\left(d-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}c-3=0\\d-4=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}c=3\\d=4\end{matrix}\right.\)

\(\Rightarrow P=25-3a-4b=25-\left(3a+4b\right)=25-Q\)

Xét \(Q=3a+4b\Rightarrow Q^2=\left(3a+4b\right)^2\le\left(3^2+4^2\right)\left(a^2+b^2\right)=25.2=50\)

\(\Rightarrow Q^2\le50\Rightarrow-5\sqrt{2}\le Q\le5\sqrt{2}\Rightarrow-Q\le5\sqrt{2}\)

\(\Rightarrow P\le25+5\sqrt{2}\)

\(P_{max}=25+5\sqrt{2}\) khi \(\left\{{}\begin{matrix}a^2+b^2=2\\\frac{a}{3}=\frac{b}{4}\\3a+4b=-5\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{3\sqrt{2}}{5}\\b=-\frac{4\sqrt{2}}{5}\end{matrix}\right.\)

\(c^2-6c+9+d^2-8d+16=0\Leftrightarrow\left(c-3\right)^2+\left(d-4\right)^2=0\Rightarrow\left\{{}\begin{matrix}c=3\\d=4\end{matrix}\right.\)

\(\Rightarrow P=25-\left(3a+4b\right)\)

Mặt khác \(\left(3a+4b\right)^2\le\left(3^2+4^2\right)\left(a^2+b^2\right)=50\)

\(\Rightarrow-5\sqrt{2}\le3a+4b\le5\sqrt{2}\)

\(\Rightarrow P\le25+5\sqrt{2}\)

\(\Rightarrow P_{max}=25+5\sqrt{2}\) khi \(\left\{{}\begin{matrix}a=\frac{-3\sqrt{2}}{5}\\b=\frac{-4\sqrt{2}}{5}\end{matrix}\right.\)

giúp mình bài này với ah.

từ hệ điều kiện, bằng cách cộng theo vế ta được:  p_^min=14 đạt được khi (2) ta nhận được 0≤b≤2⇔[b=0b=2Khi đó:-Với (2) có dạng a thỏa mãn.-Với {a^2+3c^2=28, 2a^2=2 mà ⇒{a=1c=3Vậy a=1,b=2,c=3,d=0

Từ giả thiết suy ra \(3\left(a^2+b^2+c^2+d^2\right)-d^2=42\)

\(\Leftrightarrow3Q-d^2=42\)

\(\Rightarrow Q=\dfrac{42+d^2}{3}\ge\dfrac{42}{3}=14\)

\(\Rightarrow minQ=14\Leftrightarrow\left\{{}\begin{matrix}d=0\\a^2+2b^2+3c^2=36\left(1\right)\\2a^2+b^2=6\left(2\right)\end{matrix}\right.\)

Từ \(\left(2\right)\Rightarrow b^2⋮2\Rightarrow b⋮2\)

Vì \(b^2=6-2a^2\le6\Rightarrow0\le b\le\sqrt{6}\Rightarrow b\in\left\{0;2\right\}\)

TH1: \(b=0\) ta được \(\left\{{}\begin{matrix}a^2+3c^2=36\\2a^2=6\end{matrix}\right.\Rightarrow a=\sqrt{3}\left(l\right)\)

TH2: \(b=2\) ta được \(\left\{{}\begin{matrix}a^2+3c^2=28\\2a^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=3\\a=1\end{matrix}\right.\)

Vậy \(minQ=14\Leftrightarrow\left(a;b;c;d\right)=\left(1;2;3;0\right)\)

lỗi rồi bạn nhé