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\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
P\(P=\frac{x^2+2x-4x+8}{x-4}\)
\(P=\frac{x\left(x+2\right)-4\left(x+2\right)}{x-4}\)
\(P=\frac{\left(x-4\right)\left(x+2\right)}{x-4}\)
\(P=x+2\)
2 ,\(x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)
a, \(x^4+2013x^2+2012x+2013\)
\(=x^4+2013x^2-x+2013x+2013\)
\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)
\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left\{x\left(x-1\right)+2013\right\}\)
\(=\left(x^2+x+1\right)\left(x^2-x+2013\right)\)
ĐKXĐ: \(x\ne1;3\)
\(P=\left(\frac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x^2-4}{x^2+x+1}\right)\)
\(P=\left(\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x-3}{x^2+x+1}\right)\)
\(P=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{\left(x^2+x+1\right)}{\left(x-3\right)}=\frac{x}{x-3}\)
=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]
={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]
=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]
=-x.x(x-2)(x+1)/2(2-x)x3
=(x+1)/2x
bạn Kiệt có đánh sai chỗ nào ko vậy :)). mình thấy có 1 lỗi :)).
Đặt \(a=2x+y;b=2y+x\) \(\left(a,b>0\right)\)
Khi đó : \(P=\frac{2}{\sqrt{a^3+1}-1}+\frac{2}{\sqrt{b^3+1}-1}+\frac{ab}{4}-\frac{8}{a+b}\)
Cô-si , ta có : \(\sqrt{a^3+1}=\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\frac{a+1+a^2-a+1}{2}=\frac{a^2+2}{2}\)
\(\Rightarrow\sqrt{a^3+1}-1\le\frac{a^2}{2}\)
Tương tự : \(\sqrt{b^3+1}-1\le\frac{b^2}{2}\)
Mặt khác : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Rightarrow\frac{2}{a}+\frac{2}{b}\ge\frac{8}{a+b}\Rightarrow-\frac{8}{a+b}\ge\frac{-2}{a}-\frac{2}{b}\)
\(P\ge\frac{4}{a^2}+\frac{4}{b^2}+\frac{ab}{4}-\frac{2}{a}-\frac{2}{b}=\left(\frac{4}{a^2}+1\right)+\left(\frac{4}{b^2}+1\right)+\frac{ab}{4}-\frac{2}{a}-\frac{2}{b}-2\)
\(\ge\frac{4}{a}+\frac{4}{b}+\frac{ab}{4}-\frac{2}{a}-\frac{2}{b}-2=\frac{2}{a}+\frac{2}{b}+\frac{ab}{4}-2\ge3\sqrt[3]{\frac{2}{a}.\frac{2}{b}.\frac{ab}{4}}-2=1\)
Vậy GTNN của P là 1 \(\Leftrightarrow a=b=2\Leftrightarrow x=y=\frac{2}{3}\)
Mình nghĩ đề sửa là:
Cho các số x,y nguyên. Tìm GTM của biểu thức
\(P=\frac{2}{\sqrt{\left(2x+y\right)^3+1}-1}+\frac{2}{\sqrt{\left(x+2y\right)^3+1}-1}+\frac{\left(2x+y\right)\left(x+2y\right)}{4}-\frac{8}{3\left(x+y\right)}\)
Cách làm giống @Thanh Tùng DZ@ nên không trình bày lại
a/ Ta có \(P=\frac{\frac{8}{x^2-16}+\frac{1}{x+4}}{\frac{1}{x^2-2x-8}}\)với \(\hept{\begin{cases}x\ne\pm4\\x\ne-2\end{cases}}\)
\(P=\frac{\frac{8}{\left(x-4\right)\left(x+4\right)}+\frac{1}{x+4}}{\frac{1}{\left(x-4\right)\left(x+2\right)}}\)
\(P=\frac{8+x-4}{\left(x-4\right)\left(x+4\right)}\left[\left(x-4\right)\left(x+2\right)\right]\)
\(P=x+2\)
b/ Ta có \(x^2-9x+20=20\)
<=> \(x\left(x-9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=9\end{cases}}\)
Với x = 0 thì P = x + 2 = 2
Với x = 9 thì P = x + 2 = 11
\(P=\left(\frac{8}{\left(x-4\right)\left(x+4\right)}+\frac{x-4}{\left(x+4\right)\left(x-4\right)}\right):\frac{1}{x^2-2x-8}\)
\(\Rightarrow P=\frac{x+4}{\left(x+4\right)\left(x-4\right)}:\frac{1}{x^2-2x-8}\)
\(\Rightarrow P=\frac{1}{x-4}:\frac{1}{x^2-2x-8}=\frac{x^2-2x-8}{x-4}=\frac{\left(x-4\right)\left(x+2\right)}{x-4}=x+2\)
\(b,x^2-9x+20=20\Leftrightarrow x^2-9x=0\)
\(\Rightarrow x\left(x-9\right)=0\Rightarrow\orbr{\begin{cases}x-9=0\Rightarrow x=9\Rightarrow P=9+2=11\\x=0\Rightarrow P=0+2=2\end{cases}}\)