Rút gọn biểu thức chứa căn bậc hai - Hỏi đáp

a)ĐK : a > 0

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

= \(a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

c, ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

b, ĐKXĐ: \(x>0;x\ne1\)

\(B=\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{x+1}{\sqrt{x}}\)

\(=\frac{4\sqrt{x}+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{x+1}{\sqrt{x}}\)

\(=\frac{4x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{x+1}{\sqrt{x}}=\frac{4x^2+4x}{x-1}\)

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

\(=\dfrac{\sqrt{2}.\sqrt{4-\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{2}.\sqrt{4+\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

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

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

mk chép sai đề bài ở trên nha

ĐKXĐ: ...

\(A=\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\left(\frac{\sqrt{x}-2}{\sqrt{x}+1}\right):\left(\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

Ta có: \(\left(\frac{x}{x+3\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left(1-\frac{2}{\sqrt{x}}+\frac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\frac{x+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}+\frac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{2\sqrt{x}}{\sqrt{x}+3}:\frac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

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

\(=\frac{2\sqrt{x}}{\sqrt{x}+3}\cdot\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\frac{2\sqrt{x}}{\sqrt{x}-1}\)

Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)

\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)

\(=\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}+3}{x+9}\)

\(=\frac{x+9}{\sqrt{x}-3}\cdot\frac{1}{x+9}\)

\(=\frac{1}{\sqrt{x}-3}\)

Ta có: \(P=\frac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}\)

\(=\frac{\sqrt{3}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{3}\)