Cho a,b,c >0,a+b+c=3.Chứng minh \(\dfrac{a^4}{b^2\left(c+a\right)}+\dfrac{b^4}{c^2\left(a+b\right)}+\dfrac{c^4}{a^2\left(b+c\right)}\)≥\(\dfrac{3}{2}\)
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\(3\cdot53\cdot8+4\cdot6\cdot87-212\cdot40\)
\(=24\cdot53+24\cdot87-212\cdot40\)
\(=24\cdot\left(53+87\right)-212\cdot40\)
\(=24\cdot140-212\cdot40=-5120\)
\(27\cdot38+27\cdot62\)
\(=27\left(38+62\right)=27\cdot100=2700\)
\(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
=>a=2; b=6
a+b=2+6=8
\(-5x^3+xy^2z^3\) có bậc là \(MAX\left(3;1+2+3\right)=1+2+3=6\)
Để \(\dfrac{3-x}{5}>0\) thì 3-x>0
=>x<3
=>\(x\in\left\{...;1;2;3\right\}\)
1) \(\dfrac{4^2}{2^3}=\dfrac{\left(2^2\right)^2}{2^3}=\dfrac{2^4}{2^3}=2\)
2) \(\dfrac{25^5}{125^3}=\dfrac{\left(5^2\right)^5}{\left(5^3\right)^3}=\dfrac{5^{10}}{5^9}=5\)
3) \(\dfrac{27^6}{9^9}=\dfrac{\left(3^3\right)^6}{\left(3^2\right)^9}=\dfrac{3^{18}}{3^{18}}=1\)
4) \(\dfrac{16^{13}}{32^{10}}=\dfrac{\left(2^4\right)^{13}}{\left(2^5\right)^{10}}=\dfrac{2^{52}}{2^{50}}=2^2-4\)
5) \(\dfrac{16^5}{64^4}=\dfrac{\left(4^2\right)^5}{\left(4^3\right)^4}=\dfrac{4^{10}}{4^{12}}=\dfrac{1}{4^2}=\dfrac{1}{16}\)
6) \(\dfrac{81^8}{27^{11}}=\dfrac{\left(3^4\right)^8}{\left(3^3\right)^{11}}=\dfrac{3^{32}}{3^{33}}=\dfrac{1}{3}\)
7) \(\dfrac{6^3}{2^3}=\dfrac{2^3\cdot3^3}{2^3}=3^3=27\)
8) \(\dfrac{5^4}{15^3}=\dfrac{5^4}{3^3\cdot5^3}=\dfrac{5}{3^3}=\dfrac{5}{27}\)
9) \(\dfrac{7^{15}}{14^{13}}=\dfrac{7^{15}}{7^{13}\cdot2^{13}}=\dfrac{7^2}{2^{13}}=\dfrac{49}{2^{13}}\)
10) \(\dfrac{\left(-2\right)^6}{24^2}=\dfrac{2^6}{8^2\cdot3^2}=\dfrac{2^6}{\left(2^3\right)^2\cdot3^2}=\dfrac{2^6}{2^6\cdot3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
11: \(\dfrac{27^2}{\left(-18\right)^3}=\dfrac{-3^6}{\left(3^2\cdot2\right)^3}=\dfrac{-3^6}{3^6\cdot2^3}=\dfrac{-1}{8}\)
12: \(\dfrac{\left(-10\right)^8}{8^3\cdot25^4}=\dfrac{2^8\cdot5^8}{2^6\cdot5^8}=2^2=4\)
13: \(\dfrac{4^4\cdot8^3}{16^4}=\dfrac{2^8\cdot2^9}{2^{16}}=2\)
14: \(\dfrac{5^7\cdot9^2}{15^5}=\dfrac{5^7\cdot3^4}{5^5\cdot3^5}=\dfrac{5^2}{3}=\dfrac{25}{3}\)
15: \(\dfrac{21^{13}}{49^6\cdot\left(-27\right)^4}=\dfrac{-7^{13}\cdot3^{13}}{7^{12}\cdot3^{12}}=-7\cdot3=-21\)
16: \(\dfrac{\left(-18\right)^{21}\cdot27^4}{81^{13}\cdot16^5}=\dfrac{-3^{42}\cdot2^{21}\cdot3^{12}}{3^{52}\cdot2^{20}}=\dfrac{-3^{54}}{3^{52}}\cdot2=-3^2\cdot2=-18\)
17: \(\dfrac{45^{14}\cdot8^2}{6^5\cdot125^4\cdot81^6}=\dfrac{3^{28}\cdot5^{14}\cdot2^6}{2^5\cdot3^5\cdot3^{24}\cdot5^{12}}=\dfrac{3^{28}}{3^{29}}\cdot\dfrac{5^{14}}{5^{12}}\cdot\dfrac{2^6}{2^5}=\dfrac{5^2\cdot2}{3}=\dfrac{50}{3}\)
18: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot2^2}=3\cdot\dfrac{8}{4}=3\cdot2=6\)
19: \(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{16^4\cdot5^7+20^8}\)
\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\cdot\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}=\dfrac{1}{4}\cdot5\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)
Lời giải:
Áp dụng BĐT AM-GM:
$\frac{a^4}{b^2(c+a)}+\frac{b(c+a)}{4}+\frac{b}{2}+\frac{1}{2}\geq 4\sqrt[4]{\frac{a^4}{b^2(c+a)}.\frac{b(c+a)}{4}.\frac{b}{2}.\frac{1}{2}}=2a$
Làm tương tự với các phân thức còn lại và cộng theo vế, thu gọn thì được:
$A+\frac{ab+bc+ac}{2}+\frac{a+b+c}{2}+\frac{3}{2}\geq 2(a+b+c)$
$\Leftrightarrow A\geq \frac{3}{2}(a+b+c)-\frac{3}{2}-\frac{ab+bc+ac}{2}=\frac{9}{2}-\frac{3}{2}-\frac{ab+bc+ac}{2}=3-\frac{ab+bc+ac}{2}$
Theo hệ quả quen thuộc của BĐT AM-GM:
$ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3$
$\Rightarrow A\geq 3-\frac{ab+bc+ac}{2}\geq 3-\frac{3}{2}=\frac{3}{2}$
Vậy ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$