Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)
\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)
\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y\)
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)
\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)
Dấu "=" xảy ra khi \(x=y\)
\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)
\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)
\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)
\(B_{min}=1\) khi \(x=y\)
a: ĐKXĐ: \(x,y\in R\)
b: \(A=\dfrac{\dfrac{1}{4}x^2+x^2y+\dfrac{1}{4}y+y^2+x^2y^2+\dfrac{1}{4}+\dfrac{3}{4}y}{x^2y^2+1+x^2-x^2y-y+y^2}\)
\(=\dfrac{\dfrac{1}{4}x^2+x^2y+x^2y^2+y+\dfrac{1}{4}+y^2}{x^2y^2+x^2+1+y^2-x^2y-y}\)
\(=\dfrac{\dfrac{1}{4}\left(x^2+1\right)+y\left(x^2+1\right)+x^2y^2+y^2}{\left(y^2+1\right)\left(x^2+1\right)-y\left(x^2+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+y^2}{\left(x^2+1\right)\left(y^2-y+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+y^2\left(x^2+1\right)}{\left(x^2+1\right)\left(y^2-y+1\right)}=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)
Lời giải:
Ta có: \(A=\left(1-\frac{4}{x^2}\right)\left(1-\frac{4}{y^2}\right)\)
\(A=\frac{(x^2-4)(y^2-4)}{x^2y^2}\)
\(A=\frac{[x^2-(x+y)^2][y^2-(x+y)^2]}{x^2y^2}=\frac{(-y)(2x+y)(-x)(2y+x)}{x^2y^2}\)
\(A=\frac{xy(2x+y)(2y+x)}{x^2y^2}=\frac{(2x+y)(2y+x)}{xy}=\frac{4xy+2x^2+2y^2+xy}{xy}\)
\(A=5+\frac{2(x^2+y^2)}{xy}=5+\frac{2(x-y)^2+4xy}{xy}=9+\frac{2(x-y)^2}{xy}\)
Thấy rằng \(x,y>0; (x-y)^2\geq 0\Rightarrow \frac{2(x-y)^2}{xy}\geq 0\)
\(\Rightarrow A\geq 9\) hay \(A_{\min}=9\)
Dấu bằng xảy ra khi \(x=y=1\)
\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)
Ta có \(64:59R5\Rightarrow64^n:59R5\)
Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)
Vậy \(A⋮59\)
(\(R\) là dư)
\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
cho hỏi là x=-2 thì x đâu còn \(\ge\) 0 nữa
\(\text{Khai triển ra ta được: }C=x^2y^2+2+\dfrac{1}{x^2y^2}\ge2\sqrt{\dfrac{x^2y^2}{x^2y^2}}+2=4\text{ Dấu "=" xảy ra khi: }x=\pm\dfrac{1}{y}\)