1. a) Tìm \(n\in N\)*, \(n>2008\) sao cho \(2^{2008}+2^{2012}+2^{2013}+2^{2014}+2^{2016}+2^n\) là số chính phương
b) tìm x,y > 0 thỏa mãn \(x^2+y^2=2\left(x+y\right)\left(\sqrt{x}+\sqrt{y}-2\right)\)
2. a) \(\left\{{}\begin{matrix}a\ge0\\a+b\ge1\end{matrix}\right.\). Min \(A=\frac{8a^2+b}{4a}+b^2\)
b) \(\left\{{}\begin{matrix}a,b\ge0\\\left(a-b\right)^2=a+b+2\end{matrix}\right.\). Cmr: \(\left(1+\frac{a^3}{\left(b+1\right)^3}\right)\left(1+\frac{b^3}{\left(b+1\right)^3}\right)\le9\)
c) \(x,y>0;\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=2020\). Min P = x + y
d) \(x,y,z>0;\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=6\). Min \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
e) \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z+4xyz=4\end{matrix}\right.\) Cmr: \(\left(1+xy+\frac{y}{z}\right)\left(1+yz+\frac{z}{x}\right)\left(1+zx+\frac{x}{y}\right)\ge27\)
f) \(\left\{{}\begin{matrix}x,y,z\ge1\\3x^2+4y^2+5z^2=52\end{matrix}\right.\). Min P = x + y + z
g) \(x,y>0\). Min \(P=\frac{2}{\sqrt{\left(2x+y\right)^3+1}-1}+\frac{2}{\sqrt{\left(x+2y\right)^3+1}-1}+\frac{\left(2x+y\right)\left(x+2y\right)}{4}-\frac{8}{3\left(x+y\right)}\)
?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương
giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!
Tranh thủ làm 1, 2 bài rồi ăn cơm:
1/ Đặt \(m=n-2008>0\)
\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương
\(\Rightarrow369+2^m\) là số chính phương
m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương
\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)
b/
\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)
\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)
\(\Rightarrow x=y=4\)
2/
\(A\ge\frac{8a^2+1-a}{4a}+b^2=2a+\frac{1}{4a}+b^2-\frac{1}{4}=a+\frac{1}{4a}+b^2+a-\frac{1}{4}\)
\(A\ge a+\frac{1}{4a}+b^2+1-b-\frac{1}{4}=a+\frac{1}{4a}+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}\ge1+\frac{1}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
b/ Giả thiết tương đương:
\(a\left(a+1\right)+b\left(b+1\right)=2\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow\frac{a}{b+1}+\frac{b}{a+1}=2\)
Hình như bạn ghi nhầm biểu thức
Đặt \(\left(\frac{a}{b+1};\frac{b}{a+1}\right)=\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}x+y=2\\0\le x;y\le2\end{matrix}\right.\)
\(P=\left(1+x^3\right)\left(1+y^3\right)=1+x^3+y^3+\left(xy\right)^3\)
\(=1+\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3\)
\(=\left(xy\right)^3-6xy+9=9-xy\left(6-\left(xy\right)^2\right)\)
Do \(xy\le1\Rightarrow6-\left(xy\right)^2>0\Rightarrow xy\left(6-\left(xy\right)^2\right)\ge0\)
\(\Rightarrow P\le9\Rightarrow P_{max}=9\) khi \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) hay \(\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)
Câu c giống câu này:
https://hoc24.vn/hoi-dap/question/790896.html
Bạn tham khảo tạm, cách đó quá dài nên chắc chắn ko tối ưu, nó trâu bò quá
2d/
\(P\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Đặt \(\left(\sqrt{y^2+z^2};\sqrt{z^2+x^2};\sqrt{x^2+y^2}\right)=\left(a;b;c\right)\Rightarrow a+b+c=6\)
Đổi biến được: \(\left\{{}\begin{matrix}x^2=\frac{b^2+c^2-a^2}{2}\\y^2=\frac{a^2+c^2-b^2}{2}\\z^2=\frac{a^2+b^2-c^2}{2}\end{matrix}\right.\)
\(\Rightarrow2\sqrt{2}P\ge\frac{b^2+c^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{a^2+b^2-c^2}{c}\ge\frac{\left(b+c\right)^2}{a}+\frac{\left(a+c\right)^2}{b}+\frac{\left(a+b\right)^2}{c}-6\)
\(2\sqrt{2}P\ge\frac{\left(2a+2b+2c\right)^2}{a+b+c}-6=18\Rightarrow P\ge\frac{9\sqrt{2}}{2}\)
Thực sự là câu e ko biết điểm rơi của bài toán ở đâu luôn?
f/ Đặt \(\left(x;y;z\right)=\left(a+1;b+1;c+1\right)\Rightarrow a;b;c\ge0\)
\(3\left(a+1\right)^2+4\left(b+1\right)^2+5\left(c+1\right)^2=52\)
\(\Leftrightarrow3a^2+4b^2+5c^2+6a+8b+10c=40\)
\(\Leftrightarrow5\left(a+b+c\right)^2+10\left(a+b+c\right)=40+2a^2+b^2+10\left(ab+bc+ca\right)+4a+2b\ge40\)
\(\Rightarrow\left(a+b+c\right)^2+2\left(a+b+c\right)-8\ge0\)
\(\Rightarrow\left(a+b+c+4\right)\left(a+b+c-2\right)\ge0\)
\(\Rightarrow a+b+c\ge2\Rightarrow x+y+z\ge5\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;3\right)\)
g/ Đặt \(\left\{{}\begin{matrix}2x+y=a\\x+2y=b\end{matrix}\right.\)
\(P=\frac{2}{\sqrt{a^3+1}-1}+\frac{2}{\sqrt{b^3+1}-1}+\frac{ab}{4}-\frac{8}{a+b}\)
\(=\frac{2}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}-1}+\frac{2}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}-1}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\frac{4}{a^2}+\frac{4}{b^2}+\frac{ab}{4}-\frac{8}{a+b}=2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{2}{a^2}+\frac{2}{b^2}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{4}{ab}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\frac{16}{\left(a+b\right)^2}-\frac{8}{a+b}+1-1+2\sqrt{\frac{4ab}{4ab}}\)
\(P\ge\left(\frac{4}{a+b}-1\right)^2+1\ge1\)
\(P_{min}=1\) khi \(a=b=2\Leftrightarrow x=y=\frac{2}{3}\)
g/ Đặt \(\left\{{}\begin{matrix}2x+y=a\\x+2y=b\end{matrix}\right.\)
\(P=\frac{2}{\sqrt{a^3+1}-1}+\frac{2}{\sqrt{b^3+1}-1}+\frac{ab}{4}-\frac{8}{a+b}\)
\(=\frac{2}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}-1}+\frac{2}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}-1}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\frac{4}{a^2}+\frac{4}{b^2}+\frac{ab}{4}-\frac{8}{a+b}=2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{2}{a^2}+\frac{2}{b^2}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{4}{ab}+\frac{ab}{4}-\frac{8}{a+b}\)
\(P\ge\frac{16}{\left(a+b\right)^2}-\frac{8}{a+b}+1-1+2\sqrt{\frac{4ab}{4ab}}\)
\(P\ge\left(\frac{4}{a+b}-1\right)^2+1\ge1\)
\(P_{min}=1\) khi \(a=b=2\Leftrightarrow x=y=\frac{2}{3}\)
