Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{x-\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{\sqrt{x}+2+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\frac{x-1-x+\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\frac{x+3}{\sqrt{x}+3}\)
1/
a/ \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=3+\sqrt{3}-3-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
b/ \(\sqrt{12}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
3/ \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(=\left(\dfrac{2\left(x-5\right)}{x}+\dfrac{5\left(x+10\right)}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\left(\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{10x^2-250+25x+250+x^3}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{7\left(x+5\right)^2}{5\left(x+5\right)\cdot3\left(x+5\right)}=\dfrac{7}{15}\)
3) \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right):\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x^2-25\right)+25x+250+x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{10x^2-250+25x+250-x^3}{5x\left(x+5\right)}\right).\dfrac{7}{3\left(x+5\right)}\)
\(C=\dfrac{x\left(x+2.x.5+25\right)}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x+5}{5}.\dfrac{7}{3\left(x+5\right)}=\dfrac{7}{15}\)
\(A=1-\left(\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\dfrac{2\sqrt{x}-1}{2x-x\sqrt{x}-\sqrt{x}}\)
\(=\dfrac{1-\sqrt{x}}{1-2\sqrt{x}+2x-x\sqrt{x}}\)