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Bài 1:
$14+\sqrt{40}+\sqrt{56}+\sqrt{140}=14+\sqrt{56}+(\sqrt{40}+\sqrt{140})$
=14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}=(12+2\sqrt{35})+2+(2\sqrt{10}+2\sqrt{14})$
$=(\sqrt{5}+\sqrt{7})^2+2+2\sqrt{2}(\sqrt{5}+\sqrt{7})$
$=(\sqrt{5}+\sqrt{7}+\sqrt{2})^2$
$\Rightarrow \sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{2}+\sqrt{5}+\sqrt{7}$
\(\Rightarrow A=\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=1\)
Lời giải:
a) ĐKXĐ: $a,b\geq 0$ và $a,b$ không đồng thời cùng bằng $0$
\(B=\frac{2a+2\sqrt{2}a-2\sqrt{3ab}+2\sqrt{3ab}-3b-2a\sqrt{2}}{a\sqrt{2}+\sqrt{3ab}}=\frac{2a-3b}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}=\frac{(\sqrt{2a}-\sqrt{3b})(\sqrt{2a}+\sqrt{3b})}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}\)
\(=\frac{\sqrt{2a}-\sqrt{3b}}{\sqrt{a}}=\sqrt{2}-\sqrt{\frac{3b}{a}}\)
b)
\(a=1+3\sqrt{2}; 3b=30+11\sqrt{8}\Rightarrow \frac{3b}{a}=\frac{30+11\sqrt{8}}{1+3\sqrt{2}}=\frac{(30+11\sqrt{8})(1-3\sqrt{2})}{(1+3\sqrt{2})(1-3\sqrt{2})}\)
\(=\frac{102+68\sqrt{2}}{17}=6+4\sqrt{2}=(2+\sqrt{2})^2\)
\(\Rightarrow \sqrt{\frac{3b}{a}}=2+\sqrt{2}\)
\(\Rightarrow B=\sqrt{2}-(2+\sqrt{2})=-2\)
Áp dụng BĐT Svarxơ:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\)\(=\dfrac{1}{a}+\dfrac{4}{2b}+\dfrac{9}{3c}\ge\dfrac{\left(1+2+3\right)^2}{a+2b+3c}\)\(=\dfrac{36}{a+2b+3c}\)
CMTT: \(\dfrac{2}{a}+\dfrac{3}{b}+\dfrac{1}{c}\ge\dfrac{36}{2a+3b+c}\)
\(\dfrac{3}{a}+\dfrac{1}{b}+\dfrac{2}{c}\ge\dfrac{36}{3a+b+2c}\)
Cộng vế theo vế, ta có: \(6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge36\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=36F\)
Có: \(ab+bc+ca=3abc\)
Vì a,b,c>0 nên chia cả 2 vế cho abc:
\(\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}=3\)
\(\Rightarrow36F\le18\Leftrightarrow F\le\dfrac{1}{2}\)
Vậy Fmin\(=\dfrac{1}{2}\Leftrightarrow a=b=c=1\)
Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{}\text{}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)
Ta thấy: \(\text{}\text{}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)
\(\text{}\text{}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)
Do đó: P \(\ge2+4+8=14\)
Vậy: P(min)=14 khi: \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)
Bài làm
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}=a+a+2b+b+\frac{4}{a}+\frac{9}{b}\)
\(=\left(a+2b\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)\)
\(\ge8+2\sqrt{a\times\frac{4}{a}}+2\sqrt{b\times\frac{9}{b}}\)( Cauchy )
\(=8+4+6=18\)
Đẳng thức xảy ra khi a = 2 ; b = 3
=> MinP = 18 <=> a = 2 ; b = 3
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}\)
\(\Leftrightarrow P=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)+a+2b\)
Áp dụng BĐT AM-GM ta có:
\(P\ge2.\sqrt{a.\frac{4}{a}}+2.\sqrt{b.\frac{9}{b}}+a+2b=2.2+2.3+a+2b\ge4+6+8=18\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{9}{b}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Vậy \(P_{min}=18\)\(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a}{2a+b+c}=\frac{a}{(a+b)+(a+c)}\leq \frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\frac{b}{2b+c+a}\leq \frac{b}{4}\left(\frac{1}{b+c}+\frac{1}{b+a}\right)\)
\(\frac{c}{2c+a+b}\leq \frac{c}{4}\left(\frac{1}{c+a}+\frac{1}{c+b}\right)\)
Cộng theo vế và rút gọn ta được:
\(C\leq \frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{4}\)
Vậy $C_{\max}=\frac{3}{4}$ khi $a=b=c$
\(P=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{a-b}{a\sqrt{b}+b\sqrt{a}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}:\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
Lời giải:
Trung bình cộng của 4 số là: \(\frac{50+90+a+b}{4}=70\Rightarrow a+b=140\)
Thay $a=\frac{3}{4}b$ thì:
$\frac{3}{4}b+b=140\Leftrightarrow \frac{7}{4}b=140$
$\Rightarrow b=80$
$\Rightarrw a=\frac{3}{4}b=\frac{3}{4}.80=60$
Vậy:
\(T=2a+3b=2.60+3.80=360\)
Đáp án B
Arigato :))