A = 500.500

B = 4% . 504

Vì 4% < 1 nên 4% . 504 < 504. Mà 500 . 500 > 504

=> A > B

\(a.\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(b.\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{\left(\sqrt{7}+4\right)^2}\)

\(=\left|\sqrt{7}-4\right|-\left|\sqrt{7}+4\right|\)

\(=\left(4-\sqrt{7}\right)-\left(\sqrt{7}+4\right)\)

\(=4-\sqrt{7}-\sqrt{7}-4\)

\(=-2\sqrt{7}\)

có thế n âm nhé :))

1.

\(a.\left(4x+3y\right)^2=16x^2+24xy+9y^2\)

\(b.\left(2x^3-\dfrac{1}{5}y\right)^2=4x^5-\dfrac{4}{5}x^3y+\dfrac{1}{25}y^2\)

\(c.4x^2-25y^2=\left(2x+5y\right)\left(2x-5y\right)\)

\(d.121x^2-36y^4=\left(11x+6y^2\right)\left(11x-6y^2\right)\)

2.

\(a.-6x+9x^2+1=\left(3x\right)^2-6x+1=\left(3x-1\right)^2\)

\(b.18x-x^2-81=-\left(x^2-18x+81\right)=-\left(x-9\right)^2\)

\(c.\left(x+2y\right)\left(x-2y\right)=x^2-4y^2\)

\(d.\left(5x-\dfrac{1}{3}y\right)\left(\dfrac{1}{3}y+5x\right)=\left(5x-\dfrac{1}{3}y\right)\left(5x+\dfrac{1}{3}y\right)=25x^2-\dfrac{1}{9}y^2\)

3.

\(a.96^2=\left(100-4\right)^2=10000-800+16=9216\)

\(b.103.97=\left(100+3\right)\left(100-3\right)=10000-9=9991\)

4.

\(\left(x+2y\right)\left(x-4\right)-\left(2x+1\right)\left(y-7\right)-x\left(y+5\right)\)

\(=x^2-4x+2xy-8y-2xy+14x-y+7-xy-5x\)

\(=x^2+5x-xy-9y+7\)

\(=\left(-1\right)^2+5.\left(-1\right)-2.\left(-1\right)-9.2+7=-13\)

dấu bằng thứ 2: nếu bên trong ngoặc tròn thì bên ngoài phải ngoặc vuông nhé

\(Q=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\left(x\ge0;x\ne1\right)\)

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

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

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

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

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

Thêm đk nữa nha:

\(a.x\ne-1\)

\(b.x\ne-2\)

\(e.x\ne3\)

\(f.x\ne\dfrac{1}{3}\)

\(a.\dfrac{4}{x+1}>0\)

\(\Leftrightarrow x+1>0\)

\(\Leftrightarrow x>-1\)

\(b.\dfrac{-5}{-x+2}< 0\)

\(\Leftrightarrow-x+2>0\)

\(\Leftrightarrow x< 2\)

\(c.\dfrac{x-3}{-3}< 0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(d.\dfrac{x+1}{4}\ge0\)

\(\Leftrightarrow x+1\ge0\)

\(\Leftrightarrow x\ge-1\)

\(e.\dfrac{x+2}{x-3}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\x< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)

\(f.\dfrac{2x-5}{1-3x}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\1-3x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\1-3x>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\)