\(A=\frac{1}{x-\sqrt{x}+1}=\frac{1}{x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\)

Vì\(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\Rightarrow A=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{4}{3}\)

Dấu "=" xảy ra khi \(\left(\sqrt{x}-\frac{1}{2}\right)^2=0\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

cm = quy nạp

\(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\left(\text{*}\right)\)

*Với n=1 thì (*) đúng 

*)Giả sử (*) đúng với n=k khi đó (*) thành

\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\)

Thật vậy  cm \(n=k+1\) đúng hay 

\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Lại có: \(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\frac{6\left(k+1\right)^2}{6}\)

\(=\frac{\left(k+1\right)\left[k\left(2k+1\right)+6\left(k+1\right)\right]}{6}=\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}\)

\(=\frac{\left(k+1\right)\left(2k^2+3k+4k+6\right)}{6}=\frac{\left(k+1\right)\left[\left(2k^2+3k\right)+\left(4k+6\right)\right]}{6}\)

\(=\frac{\left(k+1\right)\left[k\left(2k+3\right)+2\left(2k+3\right)\right]}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Vậy (*) đúng hay ta có DPCM

a, ĐK \(x\le2\)

\(\Rightarrow\sqrt{2-x}=2\Rightarrow2-x=4\Rightarrow x=-2\left(tm\right)\)

b, \(\sqrt{x^2-10x+25}=9\Rightarrow x^2-10x+25=81\Rightarrow x^2-10x-56=0\)

\(\Rightarrow\left(x-14\right)\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)

c. \(\sqrt{9-6x^2+x^4}=x^2+1\Rightarrow9-6x^2+x^4=x^4+2x^2+1\)do \(9-6x^2+x^4\ge0\forall x\)

\(\Rightarrow-8x^2=-8\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

\(pt< =>\hept{\begin{cases}x+y+2\sqrt{xy}=4\\x+y+6+2\sqrt{\left(x+3\right)\left(y+3\right)}=16\end{cases}}\)

<=>\(\hept{\begin{cases}x+y=4-2\sqrt{xy}\\x+y=10-2\sqrt{\left(x+3\right)\left(y+3\right)}\end{cases}}\)

=> \(4-2\sqrt{xy}=10-2\sqrt{\left(x+3\right)\left(y+3\right)}\)

<=>\(-2\sqrt{xy}=6-2\sqrt{\left(x+3\right)\left(y+3\right)}\)

<=> \(\sqrt{\left(x+3\right)\left(y+3\right)}=\sqrt{xy}+3\)

Bình phương hai vế, tự làm nốt

Lấy tổng, tích ta được:

\(\hept{\begin{cases}\sqrt{x+3}-\sqrt{x}+\sqrt{y+3}-\sqrt{y}=2\\\sqrt{x+3}+\sqrt{y}+\sqrt{y+3}+\sqrt{y}=6\end{cases}}\)Đặt \(\hept{\begin{cases}\sqrt{x+3}+\sqrt{x}=a\left(a>0\right)\\\sqrt{y+3}+\sqrt{y}=b\left(b>0\right)\end{cases}}\)và chú ý rằng \(\hept{\begin{cases}\sqrt{x+3}-\sqrt{x}=\frac{3}{a}\\\sqrt{y+3}-\sqrt{y}=\frac{3}{b}\end{cases}}\)

=>\(\hept{\begin{cases}a+b=6\\\frac{3}{a}+\frac{3}{b}=2\ge\frac{3.4}{a+b}=2\end{cases}}\)(theo Cauchy scharws)

Dấu bằng khi a=b=3

<=>x=y=1

nhan ca  tu va mau voi\(\sqrt{2}\) ta dc

\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)

                                                   =\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)

có viết đb đúng ko thế