Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)
\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)
đẳng thức khi y=-6 thủa mãn đk (*)
Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\left(x\ne\pm2\right)\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}\)
f) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
c)
\(C=4x+\frac{25}{x-1}=\left(4x-4\right)+\frac{25}{x-1}+4=4\left(x-1\right)+\frac{25}{x-1}+4\)
\(\Rightarrow C\ge2\sqrt{4\left(x-1\right).\frac{25}{x-1}}+4=20+4=24\)
Dấu "=" xảy ra khi \(4\left(x-1\right)=\frac{25}{x-1}\Leftrightarrow4\left(x-1\right)^2=25\Leftrightarrow2\left(x-1\right)=5\)( Vì \(x>1\))
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy \(Min_C=24\)
a)
\(A=x^2+xy+y^2-3x-3y+2017\)
\(\Leftrightarrow A=\left(x^2+xy+\frac{y^2}{4}\right)-3x-\frac{3}{2}y+\frac{3y^2}{4}-\frac{3}{2}y+2017\)
\(\Leftrightarrow A=\left(x+\frac{y}{2}\right)^2-2.\left(x+\frac{y}{2}\right).\frac{3}{2}+\frac{9}{4}+\left(\frac{3y^2}{4}-\frac{3}{2}y+\frac{3}{4}\right)-\frac{9}{4}-\frac{3}{4}+2017\)
\(\Leftrightarrow A=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y^2}{4}-\frac{1}{2}y+\frac{1}{4}\right)+2014\)
\(\Leftrightarrow A=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2+2014\)\(\ge2014\)\(\forall x,y\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{y}{2}-\frac{3}{2}=0\\\frac{y}{2}-\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)
Vậy \(Min_A=2014\)khi \(x=y=1\)