Mình viết nhầm bài là:

Rút gọn biểu thức:         \(A = {{12} \over√3+√2+√5}\)

Ta có: \(\left(\sqrt{12}-2\sqrt{18}+5\sqrt{3}\right)\cdot\sqrt{3}+5\sqrt{6}\)

\(=\left(2\sqrt{3}-6\sqrt{3}+5\sqrt{3}\right)\cdot\sqrt{3}+5\sqrt{6}\)

\(=3+5\sqrt{6}\)

\(12+3y+4x+xy-12+2y+6x-x-5y\)

\(=9x+xy\)

\(=x\left(y+9\right)\)

12+3y+4x+xy-12+2y+6x-x-5y=

=9x+xy

=x(9+y)

12 + 3y + 4x + xy -12 + 2y + 6x- x - 5y

=9x +xy

=x(y+9)

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

Ghi hẳn hoi đi

What ?? Đề bài j kì z bn ??

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)

a ⇒A=\(4\sqrt{4\times3}+3\sqrt{25\times3}-5\sqrt{16\times3}=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}=3\sqrt{3}\)

b ĐKXĐ x≥2 ⇔\(\sqrt{x-2}+3\sqrt{x-2}=16\Leftrightarrow4\sqrt{x-2}=16\Leftrightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Leftrightarrow x=18\)

a.   \(A=4\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)

          \(=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)

          \(=3\sqrt{3}\)

b.    \(\sqrt{x-2}-\sqrt{9x-18}=16\)

   \(\Leftrightarrow\sqrt{x-2}-\sqrt{9\left(x-2\right)}=16\)

   \(\Leftrightarrow\sqrt{x-2}-3\sqrt{x-2}=16\)

   \(\Leftrightarrow-2\sqrt{x-2}=16\)

   \(\Leftrightarrow\sqrt{x-2}=-8\)  ( Vô lý )

   Vậy PT vô nghiệm

a, 4\(\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)

=\(8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)

= (8+15-20)\(\sqrt{3}\) = 3\(\sqrt{3}\)

\(\dfrac{x}{7}=\dfrac{5}{12}\\ \Rightarrow x=\dfrac{5}{12}.7\\ \Rightarrow x=\dfrac{35}{12}\)

a) \(3x\left(x-3\right)-5x\left(x+7\right)\)

\(=3x^2-9x-5x^2-35x\)

\(=-2x^2-44x\)

b) \(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)-12\)

\(=2x^2-3x-2x^2+10x-12\)

\(=7x-12\)