Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Q=2x^2+\dfrac{6}{x^2}+3y^2+\dfrac{8}{y^2}=\left(2x^2+\dfrac{2}{x^2}\right)+\left(3y^2+\dfrac{3}{y^2}\right)+\left(\dfrac{4}{x^2}+\dfrac{5}{y^2}\right)\)
\(\ge2.2+2.3+9=19\)
Dấu = xảy ra khi \(x=y=1\)
\(Q=2x^2+\dfrac{2}{x^2}+3y^2+\dfrac{3}{y^2}+\dfrac{4}{x^2}+\dfrac{5}{y^2}\)
\(Q\ge4+6+9=19\)
###Kaito###
Ta có : a-\(\dfrac{1}{a}-2=a^2-2a+1=\left(a-1\right)^2\ge0\)
\(\Rightarrow a-\dfrac{1}{a}\ge2\)
Q(x)=2x2+\(\dfrac{2}{x^2}+3y^2+\dfrac{3}{y^2}+\dfrac{4}{x^2}+\dfrac{5}{y^2}\)
=2(\(x^2+\dfrac{1}{x^2}\)) +3(\(y^2+\dfrac{1}{y^2}\))+(\(\dfrac{4}{x^2}+\dfrac{5}{y^2}\))
\(\ge2.2+3.2+9=19\)
Dấu = xảy ra khi x=y=1
\(Q=2x^2+\frac{6}{x^2}+3y^2+\frac{8}{y^2}\)
\(=\left(2x^2+\frac{2}{x^2}\right)+\left(3y^2+\frac{3}{y^2}\right)+\left(\frac{4}{x^2}+\frac{5}{y^2}\right)\)
Ta có :
\(2x^2+\frac{2}{x^2}\ge2\sqrt{2x^2.\frac{2}{x^2}}=2\sqrt{2.2}=4\) (BĐT AM - GM)
Dấu "=" xảy ra <=> \(2x^2=\frac{2}{x^2}\Rightarrow x=1\)
\(3y^2+\frac{3}{y^2}\ge2\sqrt{3y^2.\frac{3}{y^2}}=2\sqrt{3.3}=6\) (BĐT AM - GM)
Dấu "=" xảy ra <=> \(3y^2=\frac{3}{y^2}\Rightarrow y=1\)
\(\Rightarrow Q=\left(2x^2+\frac{2}{x^2}\right)+\left(3y^2+\frac{3}{y^2}\right)+\left(\frac{4}{x^2}+\frac{5}{y^2}\right)\ge4+6+9=19\)
Dấu "=" xảy ra <=> x = y = 1
Vậỵ GTNN của Q là 19 tại x = y = 1
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\Rightarrow a+b+ab=3\)
Ta có: \(3-a+b+ab\ge ab+2\sqrt{ab}\ge3.\sqrt[3]{a^2b^2}\Leftrightarrow ab\le1\)
Suy ra \(M=\dfrac{ab}{a+1}+\dfrac{ab}{b+1}=ab.\left(\dfrac{a+1+b+1}{ab+a+b+1}\right)=ab.\dfrac{5-ab}{4}\)
\(=\dfrac{-\left[\left(ab\right)^2-2ab+1\right]+3a+1}{4}=\dfrac{-\left(ab-1\right)^2+3ab+1}{4}\le1\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=1\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
Ta có :
\(Q=\left(2x^2+\dfrac{2}{x^2}\right)+\left(3y^2+\dfrac{3}{y^2}\right)+\left(\dfrac{4}{x^2}+\dfrac{5}{y^2}\right)\ge2.2+2.3+9=19\)
Dấu "=" xảy ra khi x=y=1