a: \(\left(1-\frac{\sqrt3-1}{2}\right):\left(\frac{\sqrt3-1}{2}+2\right)\)

\(=\frac{2-\sqrt3+1}{2}:\frac{\sqrt3-1+4}{2}\)

\(=\frac{3-\sqrt3}{3+\sqrt3}=\frac{\sqrt3-1}{\sqrt3+1}=\frac{\left(\sqrt3-1\right)^2}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}\)

\(=\frac{4-2\sqrt3}{3-1}=\frac{4-2\sqrt3}{2}=2-\sqrt3\)

b: \(\frac{\sqrt5-2}{5+2\sqrt5}-\frac{1}{2+\sqrt5}+\frac{1}{\sqrt5}\)

\(=\frac{\sqrt5-2}{\sqrt5\left(\sqrt5+2\right)}-\frac{\sqrt5}{\sqrt5\left(\sqrt5+2\right)}+\frac{\sqrt5+2}{\sqrt5\left(\sqrt5+2\right)}\)

\(=\frac{\sqrt5-2-\sqrt5+\sqrt5+2}{\sqrt5\left(\sqrt5+2\right)}=\frac{\sqrt5}{\sqrt5\left(\sqrt5+2\right)}=\frac{1}{\sqrt5+2}=\sqrt5-2\)

 tham khảo

Khi p=5 thì ta có:

\(384p^2+40p+1=384\cdot5^2+40\cdot5+1\)

\(=9801=99^2\)

=>Nhận

Vậy: p=5 là giá trị cần tìm

ĐK: \(x>0;x\ne1\)

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

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

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

b) \(P=x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\) (TM)

Vậy \(P_{min}=\dfrac{3}{4}\) đạt được khi \(x=\dfrac{1}{4}\).

c) \(Q=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\) \(\Rightarrow Q>0\)

Ta có \(Q-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\)

\(\Rightarrow Q< 2\)

a) \(ĐKXĐ:x\ge0;x\ne1\)

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

\(=\dfrac{\sqrt{x}\left[\left(\sqrt{x}\right)^3-1\right]}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

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

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

b) Với \(x\ge0;x\ne1\), ta có:

\(P=x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

- Vậy \(MinP=\dfrac{3}{4}\)

c) \(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)

\(Q< 2\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}< 2\)

\(\Leftrightarrow2\sqrt{x}< 2\left(x-\sqrt{x}+1\right)\)

\(\Leftrightarrow2\sqrt{x}< 2x-2\sqrt{x}+2\)

\(\Leftrightarrow2x-4\sqrt{x}+2>0\)

\(\Leftrightarrow2\left(\sqrt{x}-1\right)^2>0\) (đúng do \(x\ne1\))

- Vậy ta có đpcm.

- Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\\sqrt{y-2}=b\ge0\\\sqrt{z-2}=c\ge0\end{matrix}\right.\)

- Đẳng thức đã cho tương đương:

\(a^2+b^2+c^2+3=2a+2b+2c\)

\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-2}=1\\\sqrt{z-2}=1\end{matrix}\right.\)

\(\Rightarrow x=y=z=3\)

\(Q=\sqrt{\left(x-y+1\right)^{2012}}+\sqrt{\left(x-3\right)^{2014}}+\sqrt{\left(y-4\right)^{2016}}\)

\(=\sqrt{1^{2012}}+\sqrt{\left(3-3\right)^{2014}}+\sqrt{\left(3-4\right)^{2016}}\)

\(=\sqrt{1}+\sqrt{0}+\sqrt{1}=2\)

Sửa đề: Chứng minh DA là tiếp tuyến của (O)

ΔOAC cân tại O

mà OH là đường cao

nên OH là phân giác của góc AOC

Xét ΔOAD và ΔOCD có

OA=OC

\(\hat{AOD}=\hat{COD}\)

OD chung

Do đó: ΔOAD=ΔOCD

=>\(\hat{OAD}=\hat{OCD}\)

=>\(\hat{OAD}=90^0\)

=>DA là tiếp tuyến của (O)

Sửa đề: \(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)

Đặt a=x-4,5

=>a-1=x-5,5

Ta có: \(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)

=>\(a^4+\left(a-1\right)^4=1\)

=>\(a^4+\left(a^2-2a+1\right)^2=1\)

=>\(a^4+a^4+4a^2+1-4a^3+2a^2-4a=1\)

=>\(2a^4-4a^3+6a^2-4a=0\)

=>\(2a\left(a^3-2a^2+3a-2\right)=0\)

=>\(a\left(a^3-a^2-a^2+a+2a-2\right)=0\)

=>\(a\left(a-1\right)\left(a^2-a+2\right)\) =0

=>a(a-1)=0

=>\(\left[\begin{array}{l}a=0\\ a-1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x-4,5=0\\ x-5,5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4,5\\ x=5,5\end{array}\right.\)

(x-4,5)⁴(x-5,5)⁴ = 0

\(\Rightarrow\left[{}\begin{matrix}x-4,5=0\\x-5,5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4,5\\x=5,5\end{matrix}\right.\)