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\(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-25\ge-25\)
\(\Rightarrow A_{min}=-25\)
\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)
\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1\right)+2093\)
\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1+3\right)+2093\)
Đặt: \(a=x^2+4x+1\)
\(\Rightarrow P=a^2-12\left(a+3\right)+2093\)
\(P=a^2-12a-36+2093\)
\(P=a^2-12a+2057\)
\(P=a^2-12a+36+2021\)
\(P=\left(a^2-2\cdot6\cdot a+6^2\right)+2021\)
\(P=\left(a-6\right)^2+2021\)
Ta có: \(\left(a-6\right)^2\ge0\forall a\)
\(\Rightarrow P=\left(t-6\right)^2+2021\ge2021\)
\(\Rightarrow P\ge2021\Rightarrow P_{min}=2021\)
Dấu "=" xảy ra: \(\left(t-6\right)^2=0\Leftrightarrow t-6=0\Leftrightarrow t=6\)
Vậy: \(P_{min}=2021\) khi \(t=6\)
Mà: \(t=6\Rightarrow x^2+4x+1=6\)
\(\Leftrightarrow x^2+4x+1-6=0\)
\(\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy: \(P_{min}=2021\) khi \(\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\\ P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+4\right)+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6-36+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6+36+2021\\ P=\left(x^2+4x-5\right)^2+2021\ge2021\)
Dấu "=" xảy ra tương đương với \(\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
ĐK: \(\left(x-2\right)\left(x^2+1\right)+2x\left(x-2\right)\ne0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\ne0\Leftrightarrow x\ne-1;2\)
Ta có: \(A=\frac{x^2\left(x-2\right)+4\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+1\right)}=\frac{x^2+4}{\left(x+1\right)^2}=\frac{t^2-2t+5}{t^2}\left(t=x+1\right)\)
\(=\frac{5}{t^2}-\frac{2}{t}+1=5\left(\frac{1}{t}-\frac{1}{5}\right)^2+\frac{4}{5}\ge\frac{4}{5}\)
Đẳng thức xảy ra khi t = 5 hay x=4
Vậy..
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt