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Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Đặt a+b=x;b+c=y;c+a=z
\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)
Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐK : \(x\ne1;x\ne2;x\ne3\)
\(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\left(\frac{x^2}{\left(x-3\right)\left(x-2\right)}+\frac{x^2}{\left(x-2\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\left(\frac{2x^2}{\left(x-1\right)\left(x-3\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\frac{2x^2}{x^4+x^2+1}\)
a, \(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-2\right)}{x^4+x^2+1}\)
\(=\left(\frac{x^2}{\left(x-3\right)\left(x-2\right)}+\frac{x^2}{\left(x-2\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\left(\frac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{x^3-x^2+x^3-3x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}.\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{2x^3-4x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}.\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{2x^3-4x^2}{\left(x-2\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2}{x^4+x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2
= 4/9 .y.y.y . (3/2-3/2.y)^2
=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)
<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5
=4/9 . 243/3125
=108/3125
Đến đó tự giải
![](https://rs.olm.vn/images/avt/0.png?1311)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
Tất cả các biểu thức này đều ko tồn tại max mà chỉ tồn tại min
\(B=\frac{x}{2}+\frac{x}{2}+\frac{4}{x^2}\ge3\sqrt[3]{\frac{4x^2}{4x^2}}=3\)
Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{4}{x^2}\Leftrightarrow x=2\)
\(C=x^2+\frac{1}{x}+\frac{1}{x}\ge3\sqrt[3]{\frac{x^2}{x^2}}=3\)
Dấu "=" xảy ra khi \(x^2=\frac{1}{x}\Leftrightarrow x=1\)
\(D=9x^2+\frac{2}{3x}+\frac{2}{3x}\ge3\sqrt[3]{\frac{36x^2}{9x^2}}=3\sqrt[3]{4}\)
Dấu "=" xảy ra khi \(9x^2=\frac{2}{3x}\Leftrightarrow x=\frac{\sqrt[3]{2}}{3}\)