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Theo mk nghĩ thì đề bài fải như thế này:
\(\left(4x^5+2x^4+4x^3-x^2-1\right):\left(2x^3+x-1\right)\)
Kết quả của phép chia trên là: \(2x^2+x+1\)
Ta có: \(2x^2+x+1=2\left(x^2+\frac{1}{2}x+\frac{1}{2}\right)\)
\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}+\frac{7}{16}\right)\)
\(=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)
=> Min = 7/8 tại \(2\left(x+\frac{1}{4}\right)^2=0\Rightarrow x=-\frac{1}{4}\)
=.= hok tốt!!
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
1) \(3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+1=3\left(x+\frac{1}{3}\right)^2+1\ge1\Rightarrow Min=1\Leftrightarrow x=-\frac{1}{3}\)
2) \(2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2-2xy+y^2+3xy\right)-3\left(x^2-2xy+y^2+4xy\right)=\left(x-y\right)^2\left(12xy-12xy\right)=0\)
3) đặt \(2x-1=t\Rightarrow x^2=\frac{t+1}{2}^2\Leftrightarrow\left(t+2\right)^3-4\frac{t+1}{2}^2\left(t-2\right)-5=0\Leftrightarrow\left(t+2\right)^3-\left(t+1\right)^2\left(t-2\right)-5=0\)\(\Leftrightarrow t^3+6t^2+12t+8-t^3-2t^2+t+2t^2+4t+2=0\Leftrightarrow6t^2+16t+10=0\Leftrightarrow\left(t+1\right)\left(6t+10\right)=0\)
=> t=-1 hoặc t=-10/6 \(\Leftrightarrow2x-1=-1\Leftrightarrow x=0\) hoặc \(2x-1=-\frac{10}{6}\Leftrightarrow x=-\frac{1}{3}\)
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
