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Bài 2:
Tìm GTLN: \(x^2+xy+y^2=3\Leftrightarrow xy=\left(x+y\right)^2-3\Rightarrow xy\ge-3\Rightarrow-7xy\le21\)
\(P=2\left(x^2+xy+y^2\right)-7xy\le2.3+21=27\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\xy=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{3},y=-\sqrt{3}\\x=-\sqrt{3},y=\sqrt{3}\end{cases}}\)
Tìm GTNN:
Chứng minh \(xy\le\frac{1}{2}\left(x^2+y^2\right)\Rightarrow\frac{3}{2}xy\le\frac{1}{2}\left(x^2+y^2+xy\right)\)
\(\Rightarrow\frac{3}{2}xy\le\frac{3}{2}\Rightarrow xy\le1\Rightarrow-7xy\ge-7\)
\(P=2\left(x^2+xy+y^2\right)-7xy\ge2.3-7=-1\)
Chúc bạn học tốt.
Làm bài 1 ha :)
Áp dụng BĐT Cô si ta có:
\(\left(1-x^3\right)+\left(1-y^3\right)+\left(1-z^3\right)\ge3\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
\(\Leftrightarrow\frac{3-\left(x^3+y^3+z^3\right)}{3}\ge\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
Mặt khác:\(\frac{3-\left(x^3+y^3+z^3\right)}{3}\le\frac{3-3xyz}{3}=1-xyz\)
Khi đó:
\(\left(1-xyz\right)^3\ge\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)\)
Giống Holder ghê vậy ta :D
Ta có : \(\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{1;-2\right\}\)
Đây giống bài lớp 6 hơn
\(4\left(x-1\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-1\right)\right]^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(x-1\right)+3\left(x+2\right)\right]\left[2\left(x-1\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(2x-2+3x+6\right)\left(2x-2-3x-6\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x+4=0\\-x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{4}{5}\\x=-8\end{array}\right.\)
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
\(\left(x-2\right)\left(x-1\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\\x^2+2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=1\\x^2=-2\left(KTM\right)\end{array}\right.\)
Vậy \(S=\left(1;2\right)\)