Đặt \(2021=a\)

\(a^2+\left(a+1\right)^2+\left(a+2\right)^2\)

\(=3a^2+6a+5\)

\(=3\left(a^2+2a+1\right)+2\)

\(=3\left(a+1\right)^2+2\) không là số chính phương

ĐK: \(\sqrt{x}>0\)

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

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

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

Chu vi hình đó là:

\(\sqrt{66,4424:3,14}\times2\times3,14=28,888cm\)

Sửa chỗ dấu bằng nha, a lm nhầm 

\(\Leftrightarrow2011-x=x-1\)

\(\Leftrightarrow x=1006\)

Vậy Min\(F=1005\sqrt{2}\Leftrightarrow x=1006\)

\(2\left(a^2+b^2\right)\ge a^2+b^2+2ab=\left(a+b\right)^2\left(\text{Cô si}\right)\)

\(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\) (Dấu "=" xảy ra⇔a=b)

Áp dụng, ta có:

\(F=\sqrt{\left(x-2011\right)^2+\left(x-1\right)^2}\)

\(=\sqrt{\left(2011-x\right)^2+\left(x-1\right)^2}\)

\(\ge\sqrt{\dfrac{1}{2}\left(2011-x+x-1\right)^2}=\sqrt{\dfrac{1}{2}\left(2010\right)^2}=\sqrt{\dfrac{2010^2}{2}}=\sqrt{\dfrac{1005^2.2^2}{2}}=1005\sqrt{2}\)

Dấu"=" xảy ra\(\Leftrightarrow x-2011=x-1\)

                      \(\Leftrightarrow0=2010\left(\text{vô lí}\right)\)

⇒ không xảy ra dấu "="

a) ĐK: \(x>1\)

b) \(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

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

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

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

\(x=\dfrac{9}{4}\) thay vào A, ta có:
\(A=\dfrac{\sqrt{\dfrac{9}{4}}}{\sqrt{\dfrac{9}{4}}-1}=\dfrac{\dfrac{3}{2}}{\dfrac{3}{2}-1}=3\)

c) để A<1

\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}< 1\)

\(\Rightarrow\sqrt{x}< \sqrt{x}-1\)

\(\Rightarrow0< -1\left(\text{vô lí}\right)\)

⇒ A>1

Ủa mà sao xóa rồi mà nó vẫn còn đây nhỉ

1. \(\dfrac{1}{5}-\dfrac{1}{2}+\left|\dfrac{-1}{4}\right|=\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{1}{4}=-\dfrac{1}{20}\)

2. \(\left|\dfrac{3}{-4}\right|+\dfrac{-1}{6}-\dfrac{5}{3}=\dfrac{3}{4}-\dfrac{1}{6}-\dfrac{5}{3}=-\dfrac{13}{12}\)

3. \(\dfrac{4}{7}-\dfrac{1}{14}+\left|\dfrac{-5}{21}\right|=\dfrac{4}{7}-\dfrac{1}{14}+\dfrac{5}{21}=\dfrac{31}{42}\)

\(x^2+2\left(m-1\right)x+m^2-m=0\)

\(\text{∆}=\left[2\left(m-1\right)\right]^2-4\left(m^2-m\right)\)

\(=4m^2-8m+4-4m^2+4m\)

\(=-4\left(m-1\right)>0\)

\(\Rightarrow m-1< 0\)\(\Rightarrow m< 1\)

Vì pt có 2 nghiệm phân biệt, áp dụng hệ thức Vi-ét, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-2m+2\\x_1.x_2=m^2-m\end{matrix}\right.\)

Ta có:

\(x_1^2+x^2_2-7x_1.x_2=10\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-9x_1.x_2=10\)

\(\Leftrightarrow\left(-2m+2\right)^2-9\left(m^2-m\right)=10\)

\(\Leftrightarrow4m^2-8m+4-9m^2+9m=10\)

\(\Leftrightarrow-5m^2+m+4=10\)

\(\Leftrightarrow-5m^2+m-6=0\)

\(\text{∆}=1^2-4.\left(-5\right).\left(-6\right)=-119< 0\)

⇒ pt vô nghiệm

a) \(x^2-5x-6=0\)

⇔\(\left(x^2+x\right)-\left(6x+6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

b) \(x^4-2x^2+1=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=0\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)