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\(A=\frac{x^9-30x^7+273x^5-820x^3+576x}{630}=\frac{x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)\left(x^2-16\right)}{630}\)
\(=\frac{\left(x-4\right)\left(x-3\right)\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{630}\)
Tử số là tích của 9 số nguyên liên tiếp nên chia hết cho cả 2; 5; 7; 9 hay chia hết 630
Vậy A luôn nguyên
\(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{9-4\sqrt{5}}}+\frac{2}{\sqrt{9+4\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{\left(\sqrt{5}-2\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=9-\frac{2}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
\(=9-\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=9-\frac{8}{5-4}\)
= 1
\(f\left(x\right)=\left(1^4-3+1\right)^{2016}=1\)
\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)
\(=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)
\(=\left(\sqrt{11}-3\right)^2\)
\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)\(\Leftrightarrow\frac{3-\sqrt{x}}{2\left(\sqrt{x}+3\right)}>0\)
vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)
\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp ĐKXĐ: \(0< x< 9\)
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
Tìm giá trị lớn nhất của \(\frac{2020-x}{6-x}\)
Ta có : \(\frac{2020-x}{6-x}=\frac{6-x+2014}{6-x}=\frac{6-x}{6-x}+\frac{2014}{6-x}=1+\frac{2014}{6-x}\)
Đa thức lớn nhất \(\Leftrightarrow1+\frac{2014}{6-x}\)lớn nhất \(\Rightarrow\frac{2014}{6-x}\)lớn nhất \(\Rightarrow6-x\)nhỏ nhất và \(6-x>0\)
Mà \(x\in Z\)\(\Rightarrow x=5\)
Vậy giá trị lớn nhất của đa thức \(=\frac{2020-5}{6-5}=2020-5=2015\)\(\Leftrightarrow x=5\)