Sẵn tiện mk chỉ cho bn luôn dạng này nhé.

Phân tích:

Với \(\alpha,\beta,\gamma>0\) thỏa \(\alpha< 2,\beta< 3,\gamma< 4\) ta có:

\(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=\left[\left(2-\alpha\right)a+\dfrac{3}{a}\right]+\left[\left(3-\beta\right)b+\dfrac{9}{2b}\right]+\left[\left(4-\gamma\right)c+\dfrac{4}{c}\right]+\left(\alpha a+\beta b+\gamma c\right)\)

\(\ge2\sqrt{3.\left(2-\alpha\right)}+2\sqrt{\dfrac{9}{2}.\left(3-\beta\right)}+2\sqrt{4.\left(4-\gamma\right)}+\left(\alpha a+\beta b+\gamma c\right)\)

Chọn \(\alpha,\beta,\gamma\) (thỏa đk trên) sao cho:

\(\left\{{}\begin{matrix}\left(2-\alpha\right)a=\dfrac{3}{a}\\\left(3-\beta\right)b=\dfrac{9}{2b}\\\left(4-\gamma\right)c=\dfrac{4}{c}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{2\left(3-\beta\right)}}\\c=\sqrt{\dfrac{4}{\left(4-\gamma\right)}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{6-4\alpha}}\\c=\sqrt{\dfrac{4}{4-3\alpha}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)

Ta có: \(a+2b+3c\ge20\). Xác định điểm rơi: \(a+2b+3c=20\)

\(\Rightarrow\sqrt{\dfrac{3}{2-\alpha}}+2\sqrt{\dfrac{9}{6-4\alpha}}+3\sqrt{\dfrac{4}{4-3\alpha}}=20\)

Giải ra ta có \(\alpha=\dfrac{5}{4}\Rightarrow\beta=\dfrac{5}{2};\gamma=\dfrac{15}{4}\)

Lời giải:

Ta có: \(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{5a}{4}+\dfrac{5b}{2}+\dfrac{15c}{4}\right)\)

\(\ge^{Cauchy}2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{5}{4}\left(a+2b+3c\right)\)

\(=3+3+2+\dfrac{5}{4}\left(a+2b+3c\right)\)

\(\ge8+\dfrac{5}{4}.20=33\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)

Vậy \(MinA=33\), đạt được khi \(a=2;b=3;c=4\)

\(đk:\left\{{}\begin{matrix}y\ne0\\y\ne\pm\end{matrix}\right.\\ =\dfrac{5}{2y\left(y+3\right)}-\dfrac{4y-3y^3}{y\left(y-3\right)\left(y+3\right)}-3\\ =\dfrac{5.\left(y-3\right)}{2y\left(y+3\right)\left(y-3\right)}-\dfrac{2\left(4y-3y^3\right)}{2y\left(y-3\right)\left(y+3\right)}-\dfrac{3.2y.\left(y^2-9\right)}{2y\left(y-3\right)\left(y+3\right)}\\ =\dfrac{5y-15-8y+6y^3-6y^3+54y}{2y\left(y^2-9\right)}\\ =\dfrac{51y-15}{2y\left(y^2-9\right)}\\ =\dfrac{3\left(17y-5\right)}{2y\left(y^2-9\right)}\)

\(c,\dfrac{5}{2y^2+6y}-\dfrac{4y-3y^3}{y^3-9y}-3\)

\(=\dfrac{5}{2y\left(y+3\right)}-\dfrac{y\left(4-3y^2\right)}{y\left(y^2-9\right)}-3\)

\(=\dfrac{5}{2y\left(y+3\right)}-\dfrac{y\left(4-3y^2\right)}{y\left(y-3\right)\left(y+3\right)}-3\)

\(=\dfrac{5\left(y-3\right)-2y\left(4-3y^2\right)-6y\left(y^2-9\right)}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\dfrac{5y-15-8y+6y^3-6y^3+54y}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\dfrac{51y-15}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\dfrac{3\left(17y-5\right)}{2y\left(y^2-9\right)}\)

1 ton of fertilizer costs: \(2975,4 : 3=991,8\)
1 pound of fertilizer costs: \(991,8 : 2,204=450\) (dollars)
 

Gọi \(x,y\) là số thứ nhất và số thứ hai

Từ đề bài, ta có hệ pt :

\(\left\{{}\begin{matrix}\dfrac{x}{y}=3\\\left(x+10\right)-\dfrac{y}{2}=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\x-\dfrac{1}{2}y=30-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\x-\dfrac{1}{2}y=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=24\\y=8\end{matrix}\right.\)

Vậy 2 số ban đầu là 24 và 8

\(2x^2-16=0\)

\(\Leftrightarrow2x^2=16\)

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow x=\pm\sqrt{8}\)

`2x^2-16=0`

`<=>2x^2=0+16`

`<=>2x^2=16`

`<=>x^2=16:2`

`<=>x^2=8`

`<=>x=+-`\(\sqrt{\text{8}}\)