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\(S=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right)}}{a+b}\)
\(\Rightarrow S.\sqrt[3]{5}=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right).5}}{a+b}\)
\(\le\dfrac{\dfrac{\left(a-2\right)+\left(b-3\right)+5}{3}}{a+b}=\dfrac{\dfrac{a+b}{3}}{a+b}=\dfrac{1}{3}\)
\(\Rightarrow S\le\dfrac{1}{3}:\sqrt[3]{5}=\dfrac{1}{3\sqrt[3]{5}}\)
Đẳng thức xảy ra \(\Leftrightarrow a-2=b-3=5\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8\end{matrix}\right.\)
Ta có BĐT \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)
Lợi dụng BĐT Cauchy-Schwarz tao cso:
\(VT^2=\left(\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+9\right)\)
\(\le3\left(\sqrt{3\left(a^2+b^2+c^2\right)}+9\right)\)
Đặt \(t=a^2+b^2+c^2\left(t\ge3\right)\) thì cần chứng minh:
\(3\left(\sqrt{3\left(a^2+b^2+c^2\right)}+9\right)\le4\left(a^2+b^2+c^2\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2+9\right)\le4\left(a^2+b^2+c^2\right)^2\)
\(\Leftrightarrow3\left(t+9\right)\le4t^2\Leftrightarrow-\left(t-3\right)\left(4t+9\right)\le0\) (Đúng)
Ta có BĐT \(3\le ab+bc+ca\le a^2+b^2+c^2\)
Và BĐT: \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)
\(\le\sqrt{9}=3\le a^2+b^2+c^2\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+9\right)\)
\(\le\left(a^2+b^2+c^2\right)\left[a^2+b^2+c^2+3\left(a^2+b^2+c^2\right)\right]\)
\(=4\left(a^2+b^2+c^2\right)=VP^2\)
Xảy ra khi \(a=b=c=1\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có: \(\sqrt{a+bc}=\sqrt{\dfrac{a^2+abc}{a}}=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\)
thiết lập tương tự ,bất đẳng thức cần chứng minh tương đương:
\(\Leftrightarrow\sum\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\ge\sqrt{abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(\Leftrightarrow\sum\sqrt{bc\left(a+b\right)\left(a+c\right)}\ge abc+\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\Leftrightarrow\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge abc+\sum a\sqrt{bc}\)
Điều này luôn đúng theo BĐT Bunyakovsky:
\(\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge\sum\left(bc+a\sqrt{bc}\right)=abc+\sum a\sqrt{bc}\)
Dấu = xảy ra khi a=b=c=3
\(\left\{{}\begin{matrix}2x-1\ge3x-9\\2-x< 2x-6\\x-3\ge4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3x\ge-9+1\\-x-2x< -6-2\\x\ge4+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\ge-8\\-3x< -8\\x\ge7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x>\dfrac{8}{3}\\x\ge7\end{matrix}\right.\Leftrightarrow7\le x\le8\)
Đề bài có vấn đề nho nhỏ, thay điểm rơi vào thì vế phải thừa bình phương trong ngoặc
Áp dụng Holder:
\(\left(a^2+\frac{1}{b^2}\right)\left(4+\frac{1}{4}\right)\left(4+\frac{1}{4}\right)\ge\left(\sqrt[3]{16a^2}+\sqrt[3]{\frac{1}{16b^2}}\right)^3\)
\(\Rightarrow\sqrt[3]{17^2\left(a^2+\frac{1}{b^2}\right)}\ge4\sqrt[3]{4a^2}+\frac{1}{\sqrt[3]{b^2}}\)
\(\Rightarrow P=\sqrt[3]{17^2}.S\ge4\sqrt[3]{4}\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}\right)+\frac{1}{\sqrt[3]{a^2}}+\frac{1}{\sqrt[3]{b^2}}+\frac{1}{\sqrt[3]{c^2}}\)
\(P=\frac{15}{\sqrt[3]{16}}\sum\sqrt[3]{a^2}+\sum\left(\sqrt[3]{\frac{a^2}{16}}+\frac{1}{\sqrt[3]{a^2}}\right)\)
Ta có: \(3\sqrt[3]{a^2}+\sqrt[3]{4}\ge4\sqrt[12]{4a^6}=4\sqrt[6]{2}.\sqrt{a}\)
Tương tự và cộng lại:
\(\Rightarrow\sum\sqrt[3]{a^2}\ge\frac{4\sqrt[6]{2}\sum\sqrt{a}-3\sqrt[3]{4}}{3}\ge3\sqrt[3]{4}\)
\(\sum\left(\sqrt[3]{\frac{a^2}{16}}+\frac{1}{\sqrt[3]{a^2}}\right)\ge6\sqrt[6]{\frac{1}{16}}=\frac{6}{\sqrt[3]{4}}\)
\(\Rightarrow P\ge\frac{15}{\sqrt[3]{16}}.3\sqrt[3]{4}+\frac{6}{\sqrt[3]{4}}=\frac{51}{\sqrt[3]{4}}=3.\sqrt[3]{\frac{17^3}{4}}\)
\(\Rightarrow S\ge3\sqrt[3]{\frac{17^3}{4}}:\sqrt[3]{17^2}=3\sqrt[3]{\frac{17}{4}}\)
Dấu "=" xảy ra khi \(a=b=c=2\)
Bài toán nhạt nhẽo, chẳng có gì ngoài tính trâu, lần sau xin né :(
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
\(P=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)
\(=\dfrac{\sqrt{\left(a-2\right).2}}{a\sqrt{2}}+\dfrac{\sqrt[3]{\left(b-3\right).\dfrac{3}{2}.\dfrac{3}{2}}}{b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{\sqrt[4]{\left(c-6\right).2.2.2}}{c\sqrt[3]{8}}\)
\(\le\dfrac{a-2+2}{2a\sqrt{2}}+\dfrac{b-3+\dfrac{3}{2}+\dfrac{3}{2}}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c-6+2+2+2}{4c\sqrt[4]{8}}\)
\(=\dfrac{a}{2a\sqrt{2}}+\dfrac{b}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c}{4c\sqrt[4]{8}}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)
Vậy \(P_{max}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-2=2\\b-3=\dfrac{3}{2}\\c-6=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=\dfrac{9}{2}\\c=8\end{matrix}\right.\)
\(P=\dfrac{bc\sqrt{a-2}+ac\sqrt[3]{b-3}+ab\sqrt[4]{c-6}}{abc}\)
\(=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)
Áp dụng BĐT AM-GM ta có:
\(=\dfrac{\sqrt{2\left(a-2\right)}}{\sqrt{2}a}+\dfrac{\sqrt[3]{2\left(b-3\right)}}{\sqrt[3]{2}b}+\dfrac{\sqrt[4]{2\left(c-6\right)}}{\sqrt[4]{2}c}\)
\(\le\dfrac{\dfrac{2+a-2}{2}}{\sqrt{2}a}+\dfrac{\dfrac{2+b-3+1}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{2+c-6+1+1+1+1}{4}}{\sqrt[4]{2}c}\)
\(=\dfrac{\dfrac{a}{2}}{\sqrt{2}a}+\dfrac{\dfrac{b}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{c}{4}}{\sqrt[4]{2}c}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{2}}+\dfrac{1}{4\sqrt[4]{2}}\)