Tìm \(\overline{abc}\), biết:
\(\hept{\begin{cases}\frac{b+c}{4}=a\\\frac{a+c}{2}=b\\a+b-1=c\end{cases}}\)
Cho\(\hept{\begin{cases}a,b,c>0\\a+b+c=1\end{cases}}\) Tìm Min\(A=\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}\)
\(A=\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\)
\(>=\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+ac+bc}\)(bđt svacxo)\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+ac+bc}+\frac{1}{ab+ac+bc}+\frac{7}{ab+ac+bc}\)
\(>=\frac{9}{a^2+b^2+c^2+ab+ac+bc+ac+ac+bc}+\frac{7}{ab+ac+bc}\)(bđt svacxo)
\(=\frac{9}{a^2+b^2+c^2+2ab+2ac+2bc}+\frac{7}{ab+ac+bc}=\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+ac+bc}\)
\(=\frac{9}{1}+\frac{7}{ab+ac+bc}=9+\frac{7}{ab+ac+bc}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc>=ab+ac+bc+2ab+2ac+2bc\)
\(=3ab+3ac+3bc=3\left(ab+ac+bc\right)\Rightarrow\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\cdot1=\frac{1}{3}>=ab+ac+bc\Rightarrow ab+ac+bc< =\frac{1}{3}\)
\(\Rightarrow9+\frac{7}{ab+ac+bc}>=9+\frac{7}{\frac{1}{3}}=9+7\cdot3=9+21=30\)
\(\Rightarrow A>=30\)dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
vậy min A là 30 khi \(a=b=c=\frac{1}{3}\)
Cho \(\hept{\begin{cases}a,b,c>0\\\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\end{cases}}=2\)
Tìm Max Q = abc
Từ đề bài \(\Rightarrow\frac{1}{1+a}=1-\frac{1}{1+b}+1-\frac{1}{1+c}=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\) (AM-GM)
Tương tự \(\hept{\begin{cases}\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(a+1\right)\left(c+1\right)}}\\\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(a+1\right)\left(b+1\right)}}\end{cases}}\)
Nhân các vế tương ứng của các bđt vừa cm đc ta có :
\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)\(\Rightarrow abc\le\frac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)
Chứng minh : \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\\a+b+c=abc\end{cases}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2}\)
Níuwqcwijnp
Tìm a, b, c thỏa mãn:
\(\hept{\begin{cases}a^4-2b=\frac{-1}{2}\\b^4-2c=\frac{-1}{2}\\c^4-2a=\frac{-1}{2}\end{cases}}\)
Tìm 2 số x; y biết rằng:
a)\(\hept{\begin{cases}\frac{x}{4}=\frac{y}{-5}\\-3x+2y=55\end{cases}}\).
b)\(\hept{\begin{cases}\frac{x}{y}=\frac{-7}{4}\\4x-5y=72\end{cases}}\).
c)\(\hept{\begin{cases}\frac{x}{-3}=\frac{y}{8}\\x^2-y^2=\frac{-44}{5}\end{cases}}\).
d)\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{-3}\\3x^3+y^3=\frac{64}{9}\end{cases}}\).
Tìm a,b,c,d >0 thỏa mãn:
\(\hept{\begin{cases}a+b+c+d=4\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4\end{cases}}\)
Ta có:
\(\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\left(a+b+c+d\right).\frac{16}{\left(a+b+c+d\right)}=16\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\)
Dấu = xảy ra khi \(a=b=c=d=1\)
Cho\(\hept{\begin{cases}a,b,c>0\\abc>1\end{cases}CMR:}2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)
Cho \(\hept{\begin{cases}a,b,c>0\\abc=1\end{cases}}\)Tìm max \(A=\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\)
Ta có \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)
Áp dụng ta có
\(a+b\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)
=> \(a+b+1\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\)
Khi đó
\(A\le\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\sqrt[3]{abc}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}=1\)
MaxA=1
Dấu bằng xảy ra khi a=b=c=1
Tìm mã của $\frac{1}{\sqrt{a^2+1}}+\frac{2}{\sqrt{b^2+4}}+\frac{3}{\sqrt{c^2+9}}$ biết \(\hept{\begin{cases}a>0,b>0,c>0\\6a+3b+2c=abc\end{cases}}\)
Thầy cô nào jup e nha.Em dốt bđt lắm
Sửa đề: Tìm Max của \(\frac{1}{\sqrt{a^2+1}}+\frac{2}{\sqrt{b^2+4}}+\frac{3}{\sqrt{c^2+9}}\) biết a,b,c>0 và 6a+3b+2c=abc