Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x-3=y\left(x+1\right)\Rightarrow y=\frac{x-3}{x+1}\)
\(A=x^2+\left(\frac{x-3}{x+1}\right)^2=x^2+\left(1-\frac{4}{x+1}\right)^2=x^2+1-\frac{8}{x+1}+\frac{16}{\left(x+1\right)^2}\)
\(=\left(x+1\right)^2-2x-\frac{8}{x+1}+\frac{16}{\left(x+1\right)^2}=\left(x+1\right)^2+\frac{16}{\left(x+1\right)^2}-2\left(x+1+\frac{4}{x+1}\right)+2\)
Đặt \(x+1+\frac{4}{x+1}=a\Rightarrow a^2=\left(x+1\right)^2+\frac{16}{\left(x+1\right)^2}+8\) (\(\left|a\right|\ge4\))
\(\Rightarrow A=a^2-8-2a+2=a^2-2a-6\)
- Nếu \(a\le-4\Rightarrow A=\left(a+4\right)^2-10a-22\ge-10a-22\ge40-22=18\)
- Nếu \(a\ge4\Rightarrow A=\left(a-4\right)^2+6a-22\ge6a-22\ge24-22=2\)
\(\Rightarrow A_{min}=2\) khi \(a=4\Rightarrow x+1+\frac{4}{x+1}=4\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Theo đề ra, ta có:
\(a^2+b^2+c^2\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
Theo BĐT Cô-si:
\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)
Ta đặt \(a^2+b^2+c^2=k\)
Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)
Vì thế nên \(k\ge\dfrac{1}{3}\)
Khi đấy:
\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)
\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).
\(B=\left(4x^2+3y\right)\left(4y^2+3x\right)+25xy\)
\(\Leftrightarrow B=16x^2y^2+12\left(x^3+y^3\right)+9xy+25xy=16x^2y^2+12\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+34xy=16t^2-2t+12\) Với t = xy
\(B=\left(4t-\dfrac{1}{4}\right)^2+\dfrac{191}{16}\)
Vì: \(0< t=xy\le\left(\dfrac{x+y}{2}\right)^2=\dfrac{1}{4}\Rightarrow\dfrac{-1}{4}< 4t-\dfrac{1}{4}\le\dfrac{3}{4}\)
Vậy \(\dfrac{191}{16}\le B\le\dfrac{25}{2}\)
Để phương trình có nghiệm thì :
\(\Delta_x=a^2-\left(2a^2+b^2-5\right)\ge0\)
\(\Leftrightarrow a^2+b^2\le5\)
\(\Leftrightarrow\left(a+b\right)^2\le5+2ab\)
\(\Leftrightarrow ab\ge\frac{\left(a+b\right)^2-5}{2}\)
Ta có :
\(P=\left(a+1\right)\left(b+1\right)=ab+a+b+1\)
\(\ge\frac{\left(a+b\right)^2-5}{2}+\left(a+b\right)+1=\frac{1}{2}\left(a+b+1\right)^2-2\ge-2\)
Dấu " = " xảy ra khi \(\hept{\begin{cases}a=-2\\b=1\end{cases}}\)
\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow\left(x+y\right)^2\le2.1=2\Rightarrow-\sqrt{2}\le\left(x+y\right)=A\le\sqrt{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{2}{\left(x+y\right)^2}=2\left(x+y=1\right)\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)
\(GT\Rightarrow\)\(\frac{1}{a+2}+\frac{3}{b+4}\leq1-\frac{2}{c+3}\)
Áp dụng BĐT AM-GM ta có:
\(1-\frac{2}{c+3}\geq\frac{1}{a+2}+\frac{3}{b+4}\geq2\sqrt{\frac{3}{(a+2)(b+4)}}\)
Tương tự ta có:
\(1-\frac{1}{a+2}\geq\frac{3}{b+4}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(b+4)}}\)
\(1-\frac{3}{b+4}\geq\frac{1}{a+2}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(a+2)}}\)
Nhân theo vế ta được: \((1-\frac{2}{c+3})(1-\frac{1}{a+2})(1-\frac{3}{b+4})\geq \frac{48}{(a+2)(b+4)(c+3)}\)
\(\Leftrightarrow (\frac{c+1}{c+3})(\frac{a+1}{a+2})(\frac{b+1}{b+4})\geq\frac{48}{(a+2)(b+4)(c+3)}\)
\(\Leftrightarrow(a+1)(b+1)(c+1)\geq48\)
Dấu "=" xảy ra khi \(a=1;c=3;b=5\)
\(Gt\Leftrightarrow 1-\frac{1}{a+2}+1-\frac{3}{b+4}+\frac{c+1}{c+3}\geq 2\\\Leftrightarrow \frac{a+1}{a+2}+\frac{b+1}{b+4}+\frac{c+1}{c+3}\geq 2\)
Đặt \((a+1;b+1;c+1)\rightarrow (x;y;z)\), vậy cần tìm GTNN của \(Q=xyz\)
Ta có: \(\frac{x}{x+1}+\frac{y}{y+3}+\frac{z}{z+2}\geq 2\)
Áp dụng BĐT AM-GM ta có:
\(\frac{x}{x+1}\geq 1-\frac{y}{y+3}+1-\frac{z}{z+2}=\frac{3}{y+3}+\frac{2}{z+2}\geq 2\sqrt{\frac{6}{(y+3)(z+2)}}\)
\(\frac{y}{y+3}\geq 1-\frac{x}{x+1}+1-\frac{z}{z+2}=\frac{1}{x+1}+\frac{2}{z+2}\geq 2\sqrt{\frac{2}{(x+1)(z+2)}}\)
\(\frac{z}{z+2}\geq 1-\frac{x}{x+1}+1-\frac{y}{y+3}= \frac{1}{x+1}+\frac{3}{y+3}\geq 2\sqrt{\frac{3}{(x+1)(y+3)}}\)
Nhân theo vế ta có:\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{48}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\Leftrightarrow Q\ge48\)
Dấu "=" xảy ra khi \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{x+1}=\frac{3}{y+3}=\frac{2}{z+2} & & \\ \frac{1}{a+2}+\frac{3}{b+4}=\frac{c+1}{c+3} & & \end{matrix}\right.\)\(\Leftrightarrow a=1;b=5;c=3\)
Để \(Q\) nhỏ nhất => \(m;n\) nhỏ nhất
=>\(m^2+n^2\) nhỏ nhất
Mà \(m^2;n^2\ge0\)
Suy ra để \(Q\) nhỏ nhất thì
\(m=n=0\) thay \(m=0;n=0\) vào \(Q\) đc kq