\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

Rút gọn các biểu thức sau với x≥0x≥0:

a) 2\(\sqrt{3x}\)-4\(\sqrt{3x}\)+27-3\(\sqrt{3x}\)=27-5\(\sqrt{3x}\)

b)3\(\sqrt{2x}\)-5\(\sqrt{8x}\)+7\(\sqrt{18x}\)+28

=3\(\sqrt{2x}\)-10\(\sqrt{2x}\)+21\(\sqrt{2x}\)+28

=14\(\sqrt{2x}\)+28=14(\(\sqrt{2x}\)+2)

a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)

\(=\left(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}\right)+27\)

\(=-5\sqrt{3x}+27\)

b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)

\(=3\sqrt{2x}-5\sqrt{4.2x}+7\sqrt{9.2x}+28\)

\(=3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}+28\)

\(=3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}+28\)

\(=\left(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}\right)+28\)

\(=\left(3-10+21\right)\sqrt{2x}+28\)

\(=14\sqrt{2x}+28\)

\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{3x+8\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{20\sqrt{x}+6x-10}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

Đề sai không vậy?

Ta có: \(N=\dfrac{3x+5\sqrt{x}}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{7}{\sqrt{x}+3}\)

\(=\dfrac{3x+5\sqrt{x}+3x+9\sqrt{x}-\sqrt{x}-3+7\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{6x+20\sqrt{x}-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)

b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)

\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\) 

\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)

a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)

b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\) 

\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)

a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)

\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi x=1