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\(=\frac{\left(x^2+1\right)\left(x^8+2x^4+1-x^4\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)}{x^4+x^2+1}\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
\(\frac{\left(x^2+1\right)\left(x^8+x^4+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{\left(x^2+1\right)\left(x^8+2x^4+1-x^4\right)}{\left(x^2+1\right)^2-x^2}\)
\(=\frac{\left(x^2+1\right)\left[\left(x^4+1\right)^2-x^4\right]}{x^4+2x^2+1-x^2}=\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)}{x^4+x^2+1}\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
Ta có: \(5x^2+10yz\le5\left(x^2+y^2+z^2\right)=9x\left(y+z\right)+18yz\)\(\Leftrightarrow5x^2\le9x\left(y+z\right)+8yz\le9x\left(y+z\right)+2\left(y+z\right)^2\)\(\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9\left(\frac{x}{y+z}\right)-2\le0\Leftrightarrow\left(\frac{5x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\)
\(\Rightarrow\frac{x}{y+z}\le2\)(Do \(\frac{5x}{y+z}+1>0\forall x,y,z>0\))
\(\Leftrightarrow x\le2\left(y+z\right)\Leftrightarrow x+y+z\le3\left(y+z\right)\)
\(\Rightarrow P\le\frac{2x}{\left(y+z\right)^2}-\frac{1}{\left(x+y+z\right)^3}\le\frac{4\left(y+z\right)}{\left(y+z\right)^2}-\frac{1}{\left(3y+3z\right)^3}\)
\(=\frac{4}{y+z}-\frac{1}{27\left(y+z\right)^3}\)
Đặt \(\frac{1}{y+z}=t\)thì \(P\le4t-\frac{1}{27}t^3-16+16=-\frac{1}{27}\left(t-6\right)^2\left(t+12\right)+16\le16\)
Vậy MaxP = 16 khi \(\left(x,y,z\right)=\left(\frac{1}{3},\frac{1}{12},\frac{1}{12}\right)\)
Gọi x là dộ dài quãng đường ab \(x\ge0\) ( km )
Thời gian nếu đi như dự định \(\frac{x}{40}\)
Thời gian đi lúc đầu \(\frac{\frac{1}{2}x-60}{40}=\frac{x-120}{80}\)
Vận tốc lúc sau 40 + 10 = 50
Thời gian đi lúc sau \(\frac{\frac{1}{2}x+60}{50}=\frac{x+120}{100}\)
Theo đề , ta có
\(\frac{x-120}{80}+\frac{x+120}{100}=\frac{x}{40}-1\)
\(\frac{5x-600}{400}+\frac{4x+480}{400}=\frac{10x}{400}-\frac{400}{400}\)
\(5x-600+4x+480=10x-400\)
\(9x-120=10x-400\)
\(400-120=10x-9x\)
\(x=280\)
Vậy quãng đường AB dài 280 km
MTC : ( x - 1 )( x2 + x + 1 )
Ta có : \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
Hnay mới học thì hnay trả lời nhá :P
\(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1}\)
Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2+x+1=x^2+x+1\)
MTC : \(\left(x-1\right)\left(x^2+x+1\right)\)
\(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(2x\left(x-7\right)+7-x=0\Leftrightarrow2x\left(x-7\right)-\left(x-7\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-7\right)=0\Leftrightarrow x=\frac{1}{2};7\)
a, \(5x-15y=5\left(x-3y\right)\)
b, \(12y\left(2x-5y\right)+6xy\left(5-2x\right)=12y\left(2x-5\right)-6xy\left(2x-5\right)\)
\(=6y\left(2-x\right)\left(2x-5\right)\)
c, \(x^2-7x+12=x^2-3x-4x+12=\left(x-4\right)\left(x-3\right)\)