\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3

a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

a, A=2(-1/4)x⁵y³.

-Bậc=8.

a,P(\(x\)) =  \(x^3\) - 2\(x\) + 6 + 3\(x\)⁴ - \(x\) + 2\(x\)³ - 2\(x\)²

   P(\(x\)) = (\(x^3\) + 2\(x^3\)) - ( 2\(x\) + \(x\) ) + 6 + 3\(x^4\) - 2\(x^2\)

   P(\(x\))  = 3\(x^3\) - 3\(x\) + 6 + 3\(x^4\)- 2\(x^2\)

   P(\(x\) )= 3\(x^4\) + 3\(x^3\) - 2\(x^2\) - 3\(x\) + 6

    Q(\(x\)) = \(x^3\) -  7 + 2\(x^2\) + 3\(x\) - 9\(x^2\) - 2 - 4\(x^3\)

   Q(\(x\)) =  (\(x^3\) - 4\(x^3\)) - ( 7 + 2) - (9\(x^2\) - 2\(x^2\)) + 3\(x\)

   Q(\(x\)) = -3\(x^3\) - 9 - 7\(x^2\) + 3\(x\)

  Q(\(x\)) = -3\(x^3\) - 7\(x^2\) + 3\(x\) - 9

Bậc  cao nhất của P(\(x\)) là 4; hệ số cao nhất là: 3; hệ số tự do là 6

Bậc cao nhất của Q(\(x\)) là 3; hệ số cao nhất là -3; hệ số tự do là -9

Bài 1:

a)  \(B=\sqrt{1-4x+4x^2}\)

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

         \(=\left|1-2x\right|\)

Nếu  \(x\le\frac{1}{2}\)thì:  \(B=1-2x\)

Nếu  \(x>\frac{1}{2}\)thì:  \(B=2x-1\)

b)  Tại \(x=-7\)thì:  \(B=1-2.\left(-7\right)=15\)

Bài 2:

\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.2+2^2}+\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\sqrt{3}+2+2-\sqrt{3}=4\) (đpcm)