Giải:

Ta có:

\(P=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)

\(=\left[\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right].\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{x-\sqrt{x}+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)

b) Giải:

Ta có: \(k\left(k+1\right)\left(k+2\right)\)

\(=\dfrac{1}{4}\left[k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\left(k-1\right)k\left(k+1\right)\left(k+2\right)\right]\)

Do đó: \(P=\dfrac{1}{4}.n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

Thay vào ta tính được:

\(P\left(100\right)=26527650;P\left(2009\right)=\dfrac{1}{4}.2009.2010.2011.2012\)

Mà: \(\dfrac{1}{4}.2009.2010.2011=2030149748\)

Và \(149748.2012=3011731776;2030.2012.10^6=4084360000000\)

Cộng lại ta có: \(P\left(2009\right)=4087371731776\)

Giải:

Ta có: \(U_{n-1}=\dfrac{3U_n-U_{n+1}}{2}\) nên:

\(U_4=340;U_3=216;U_2=154;U_1=123\)

Từ \(U_5=588;U_6=1084;U_{n+1}=3U_n-2U_{n-1}\)

\(\Rightarrow\) \(U_{25}=520093788\)

Vậy \(U_2=154;U_1=123;\) \(U_{25}=520093788\)

a) 60

b) 792

c) ukmmmmmmmmmm chưa ra

Giải:

Ta có:

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\left(1\right)\)

\(\Rightarrow3A=3\left(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\right)\)

\(=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\left(2\right)\)

Lấy \(\left(2\right)+\left(1\right)\Leftrightarrow4A=3^{101}+1\)

\(\Rightarrow A=\dfrac{3^{101}+1}{4}\)

Giải:

\(\dfrac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}\) \(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)

\(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)

Áp dụng vào biểu thức ta có:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}\) \(+...+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)

\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}\)

\(=1-\dfrac{1}{\sqrt{2015}}\)

Giải:

Từ \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\)

\(\Leftrightarrow\left(a+c\right)^3=-\left(b+d\right)^3\)

\(\Leftrightarrow a^3+c^3+3ac\left(a+c\right)=-\left[b^3+d^3+3bd\left(b+d\right)\right]\)

\(\Leftrightarrow VT=a^3+b^3+c^3+d^3=-3bd\left(b+d\right)-3ac\left(a+c\right)\)

\(=-3bd\left(b+d\right)+3ac\left(b+d\right)=3\left(ac-bd\right)\left(b+d\right)=VP\) (Đpcm)

Giải:

Ta có:

\(E=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-6^2\ge-6^2=-36\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x^2+5x\right)^2\ge0\Leftrightarrow\) \(\left[\begin{array}{}x=0\\x=-5\end{array}\right.\)

Vậy \(x=0\) hoặc \(x=-5\) thì \(E_{min}=-36\)