Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)
c) \(A^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2y^2+x^2+x^2y^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2-1=2010-1=2009\)
Vì A>0 nên \(A=\sqrt{2009}\)
d) \(2009^2=\left(2008+1\right)^2=2008^2+2.2008+1\)
\(1+2008^2=2009^2-2.2008=2009^2-2.2009\dfrac{2008}{2009}\)
\(A=\sqrt{2009^2-2.2009.\dfrac{2008}{2009}+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)
\(A=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
Câu 1:
\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)
\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)
Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)
\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)
Và do $a,b\geq 3$ nên:
\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)
\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)
Cộng tất cả những BĐT trên ta có:
\(A\geq 2+14+62+2=80\) (đpcm)
Dấu "=" xảy ra khi $a=b=3$
Câu 2:
Bình phương 2 vế ta thu được:
\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)
\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)
\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)
\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)
Do đó pt đã cho vô nghiệm.
2
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)