Đặt 2011=t

\(\Rightarrow T=\sqrt{1+\left(t-1\right)^2+\frac{\left(t-1\right)^2}{t^2}}+\frac{t-1}{t}\)

        \(=\sqrt{\frac{t^2+t^2\left(t-1\right)^2+\left(t-1\right)^2}{t^2}}+\frac{t-1}{t}\)

        \(=\frac{\sqrt{t^2+t^4-2t^3+t^2+t^2-2t+1}+t-1}{t}\)

        \(=\frac{\sqrt{t^4+t^2+1+2t^2-2t^3-2t}+t-1}{t}\)

         \(=\frac{\sqrt{\left(t^2-t+1\right)^2}+t-1}{t}\)

       \(=\frac{t^2-t+1+t-1}{t}=t=2011\)

mà \(2011\in Z\)

nên T là một số nguyên.

THAM KHẢO CÁCH RÚT GỌN DÒNG 3: Chứng minh đẳng thức lượng giác sin^4a + cos^4a = 1 - 1/2 sin^2 2a = 3/4 + 1/4 cos4a câu hỏi 1063664 - hoidap247.com

\(M=\sin^4\alpha\left(1+2\cos^2\alpha\right)+\cos^4\alpha\left(1+2\sin^2\alpha\right)\\ M=\left(\sin^4\alpha+\cos^4\alpha\right)+2\sin^2\alpha\cdot\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)\\ M=\dfrac{3}{4}+\dfrac{1}{4}\cos4\alpha+\dfrac{1}{2}\sin^22\alpha\cdot1\\ M=\dfrac{3}{4}+\dfrac{1}{4}\cos4\alpha+\dfrac{1}{2}\sin^22\alpha\)

a) Thay x=4(TMĐK) vào B ta có: 

\(B=\dfrac{4-\sqrt{4}}{2\sqrt{4}+1}=\dfrac{2}{5}\)

Vậy x=4 thì B=\(\dfrac{2}{5}\)

b)\(M=A.B\)

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

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

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

M= \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

c)\(M=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+1\)

\(\Leftrightarrow2\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy với x=\(\dfrac{1}{4}\) thì M=\(\dfrac{1}{3}\)

`a)x=64`

`=>N=sqrtx/(sqrtx-3)=8/(8-3)=8/5`

`b)M=(2sqrtx)/(sqrtx-3)-(x+9sqrtx)/(x-9)`

`=(2x+6sqrtx-x-9sqrtx)/(x-9)`

`=(x-3sqrtx)/(x-9)`

`=sqrtx/(sqrtx+3)`

`P=M.N=x/(x-9)`

`c)` So sánh gì với 1?

a) Thay x=64(TMĐK) vào N ta có: 

\(N=\dfrac{\sqrt{64}}{\sqrt{64}-5}=\dfrac{8}{3}\)

Vậy x=64 thì N=\(\dfrac{8}{3}\)

b) \(P=M.N\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x-9\sqrt{x}}{x-9}.\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}\right)\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}\right)\)

\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{\sqrt{x}+5}\)

\(P=\dfrac{x+15\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{\sqrt{x}+5}\)

\(P=\dfrac{x}{\left(\sqrt{x}-3\right)\left(\sqrt{x} +3\right)}\)

\(P=\dfrac{x}{x-9}\)

\(\Leftrightarrow\dfrac{2m+3}{5}=\dfrac{5m+2}{3}\\ \Leftrightarrow6m+9=25m+10\\ \Leftrightarrow19m=-1\Leftrightarrow m=-\dfrac{1}{19}\)

mik cảm ơn bn nhìu nha

\(1,A=10\sqrt{2}+5\sqrt{2}-6\sqrt{2}=9\sqrt{2}\\ B=6\sqrt{3}-4\sqrt{3}-\sqrt{3}=\sqrt{3}\\ 2,\\ a,ĐK:1-3x\ge0\Leftrightarrow x\le\dfrac{1}{3}\\ b,ĐK:x\ge0;x\ne4\\ 3,\\ a,\Leftrightarrow12x-3=4\Leftrightarrow x=\dfrac{7}{12}\\ b,\Leftrightarrow\left|2x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow\left|3x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\1-3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(4,\\ B=\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{4\left(a+1\right)}\\ B=\dfrac{2\left(a+1\right)}{4\left(a+1\right)}=\dfrac{1}{2}\)