Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2. voi a1,a2,a3 duong nhân từng vế của hai phương trình\(\left(a_1+a_2+a_3\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)=9\)
áp dụng phương pháp bdt không chặt thì pt trên xảy ra <=>\(a_1=a_2=a_3=1\)
1.
tu pt 2 ta co
dk: y(y+1) khac 0
x(x+1)=72/y(y+1)
the vao 1 ta co
\(\frac{72}{y\left(y+1\right)}+y\left(y+1\right)=18\)
<=>\(y^2\left(y+1\right)^2-18y\left(y+1\right)+81-9=0\)
<=>\(\left[y\left(y+1\right)-9\right]^2=3\)
tu giai tiep
3, \(\sqrt{\frac{a}{b+c}}=\sqrt{\frac{a^2}{a\left(b+c\right)}}\Rightarrow\frac{1}{\sqrt{\frac{a}{b+c}}}=\sqrt{\frac{a\left(b+c\right)}{a^2}}.\)
Áp dụng bất đẳng thức Cô si ta có : \(\sqrt{\frac{a\left(b+c\right)}{a^2}}\le\frac{a+b+c}{2a}\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\left(1\right).\)
Chứng minh tương tự ta có : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\left(2\right).\); \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\left(3\right).\)
Cộng vế với vế các bất đẳng thức cùng chiều ta được:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2.\)( đpcm )
dấu " = " xẩy ra khi a = b = c > 0
\(\frac{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}{2\sqrt{3}}=\frac{A+B\sqrt{3}-A+B\sqrt{3}}{2\sqrt{3}}=B\)( A,B thuộc Z )
\(\sqrt{\left(\sqrt{3}-2\right)^2\cdot a^2}\)
\(=\left|a\cdot\left(2-\sqrt{3}\right)\right|\)
\(=\left(2-\sqrt{3}\right)\cdot\left|a\right|\)
1, A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
ĐKXĐ: a≥0
A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+1\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1}{a+1}-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right).\left(\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}\right)\)
A=\(\sqrt{a}+1\)
Vậy A=\(\sqrt{a}+1\)
2, a=1996-2\(\sqrt{1995}\)
a=\(1995-2\sqrt{1995}+1\)
a=\(\left(\sqrt{1995}-1\right)^2\) (TMĐKXĐ)
thay a=\(\left(\sqrt{1995}-1\right)^2\) vào A ta có:
A=\(\sqrt{\left(\sqrt{1995}-1\right)^2}+1\)
A=\(\sqrt{1995}\)
Vậy a=1996-2\(\sqrt{1995}\) thì A=\(\sqrt{1995}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$C^2\leq (a+b)[(29a+3b)+(29b+3a)]=32(a+b)^2$
$(a+b)^2\leq (a^2+b^2)(1+1)\leq 4$
$\Rightarrow C^2\leq 32.4$
$\Rightarrow C\leq 8\sqrt{2}$
Vậy $C_{\max}=8\sqrt{2}$. Dấu "=" xảy ra khi $a=b=1$
\(P=2+\dfrac{2}{b}+a+\dfrac{a}{b}+2+\dfrac{2}{a}+b+\dfrac{b}{a}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\left(\dfrac{3}{2a}+\dfrac{3}{2b}\right)+4\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{a.\dfrac{1}{2a}}+2\sqrt{b.\dfrac{1}{2b}}+2\sqrt{\dfrac{3}{2a}.\dfrac{3}{2b}}+4=6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\)
Ta lại có: \(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\left(BĐT.Cauchy\right)\Rightarrow2\left(a^2+b^2\right)\ge4ab\Rightarrow\sqrt{ab}\le\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow P\ge6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\ge6+2\sqrt{2}+\dfrac{3}{\dfrac{\sqrt{2}}{2}}=6+5\sqrt{2}\)
\(minP=6+5\sqrt{2}\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)