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a)ĐKXĐ:\(a\ge0;a\ne16\)
\(B=\left[\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right]:\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
=\(\dfrac{3\sqrt{a}\left(\sqrt{a}-4\right)+\sqrt{a}\left(\sqrt{a}+4\right)-4\left(a+2\right)}{a-16}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}=\dfrac{-8\sqrt{a}-8}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8\left(-\sqrt{a}-1\right)}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8}{\sqrt{a}-4}\)
Vậy...
b)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
B=-3 thì \(\dfrac{8}{\sqrt{a}-4}=-3\)
=>\(9=-3\sqrt{a}+24\)
<=>-15=-3\(\sqrt{a}\)
<=>\(\sqrt{a}=5\)
<=>a=25(TM)
Vậy a=25 thì B=-3
c)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
| \(\sqrt{a}-4\) | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| \(\sqrt{a}\) | -4(L) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| \(\sqrt{a}\) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| a | 0(TM) | 4(TM) | 9(TM) | 25(TM) | 36(TM) | 64(TM) | 144(TM) |
(BẠN KẺ 1 BẢNG 3 HÀNG THÔI NHA,MÌNH KẺ LỖI NÊN LÀM 2 BẢNG)
Vậy...
Theo BĐT Cauchy: \(\left\{{}\begin{matrix}\sqrt{ab}\le\dfrac{a}{4}+b\\\sqrt[3]{abc}\le\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)\end{matrix}\right.\)
\(\Rightarrow a+\sqrt{ab}+\sqrt[3]{abc}\le a+\dfrac{a}{4}+b+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)=\dfrac{4}{3}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1346}{\dfrac{4}{3}\left(a+b+c\right)}-\dfrac{2019}{\sqrt{a+b+c}}=\dfrac{2019}{2\left(a+b+c\right)}-\dfrac{2019}{\sqrt{a+b+c}}\)
\(\Rightarrow\dfrac{2P}{2019}\ge\dfrac{1}{a+b+c}-\dfrac{2}{\sqrt{a+b+c}}=\left(\dfrac{1}{\sqrt{a+b+c}}\right)^2-2.\dfrac{1}{\sqrt{a+b+c}}+1-1\)
\(\Rightarrow\dfrac{2P}{2019}\ge\left(\dfrac{1}{\sqrt{a+b+c}}-1\right)^2-1\ge-1\)
\(\Rightarrow P\ge\dfrac{-2019}{2}\)
\(\Rightarrow P_{min}=\dfrac{-2019}{2}\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\dfrac{a}{4}=b=4c\\\dfrac{1}{\sqrt{a+b+c}}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{16}{21}\\b=\dfrac{4}{21}\\c=\dfrac{1}{21}\end{matrix}\right.\)
a) Gọi q là công sai của cấp số nhân. Ta có: \(a;b=aq;c=aq^2\).
\(a^2b^2c^2\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{b^2c^2}{a}+\dfrac{a^2c^2}{b}+\dfrac{a^2b^2}{c}\)
\(=\dfrac{\left(a.q\right)^2\left(a.q^2\right)^2}{a}+\dfrac{a^2\left(aq^2\right)^2}{aq}+\dfrac{a^2\left(aq\right)^2}{aq^2}\)
\(=\dfrac{a^2q^2a^2q^4}{a}+\dfrac{a^2a^2q^4}{aq}+\dfrac{a^2a^2q^2}{aq^2}\)
\(=a^3q^6+a^3q^3+a^3\)
\(=\left(a^2q\right)^3+\left(aq\right)^3+a^3\)
\(=c^3+b^3+a^3=a^3+b^3+c^3\).
b) Gọi q là công bội của của cấp số nhân.
Ta có: \(a;b=aq;c=aq^2;d=aq^3\).
\(\left(ab+bc+cd\right)^2=\left(a.aq+aq.aq^2+aq^2.aq^3\right)^2\)
\(=\left(a^2q+a^2q^3+a^2q^5\right)^2=a^4q^2\left(1+q^2+q^4\right)^2\). (1)
\(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)\)\(=\left(a^2+a^2q^2+a^2q^4\right)\left(a^2q^2+a^2q^4+a^2q^6\right)\)
\(=a^2\left(1+q^2+q^4\right)a^2q^2\left(1+q^2+q^4\right)\)
\(=a^4q^2\left(1+q^2+q^4\right)^2\). (2)
So sánh (1) và (2) ta có điều phải chứng minh.
1: \(Q=\dfrac{ab\left(a-b\right)}{ab}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
2: \(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-...-\sqrt{2001}+\sqrt{2005}}{4}\)
\(=\dfrac{\sqrt{2005}-1}{4}\)
Áp dụng BĐT Holder:
\(\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\ge\left(a^2+b^2+c^2\right)^3\)
Mặt khác:
\(\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\dfrac{3}{2}\left(a^2b^2+b^2c^2+c^2a^2+abc\left(a+b+c\right)\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{3}{4}\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\)
\(\Rightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\)
\(\Rightarrow P\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}+\dfrac{4}{\sqrt{a^2+b^2+c^2+1}}\)
Đặt \(\sqrt{\dfrac{a^2+b^2+c^2}{3}}=x>0\)
\(\Rightarrow P\ge\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\)
Ta sẽ chứng minh \(P\ge\dfrac{7}{2}\)
Thật vậy, với \(x\ge\dfrac{7}{3}\Rightarrow P>\dfrac{3x}{2}\ge\dfrac{7}{2}\) (đúng)
Với \(0< x\le\dfrac{7}{3}\) ta cần chứng minh:
\(\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7}{2}\Leftrightarrow\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7-3x}{2}\)
\(\Leftrightarrow64\ge\left(7-3x\right)^2\left(3x^2+1\right)\)
\(\Leftrightarrow3\left(x-1\right)^2\left(-9x^2+24x+5\right)\ge0\)
\(\Leftrightarrow\left(x-1\right)^2\left[3x\left(7-3x\right)+3x+5\right]\ge0\) (đúng)
Vậy \(P_{min}=\dfrac{7}{2}\) khi \(x=1\) hay \(a=b=c=1\)