mình mới học lớp 6 nên không biết

mày xem sẽ chưa tao xem rồi đó về mà xem

mất dạy thì biến nha

a) \(\sqrt{\frac{2}{3}}=\sqrt{\frac{2.3}{3.3}}=\frac{\sqrt{6}}{3}\)

b) \(\sqrt{\frac{x^2}{2}}=\sqrt{\frac{2x^2}{2.2}}=\frac{x\sqrt{2}}{2}\)

c) \(\sqrt{\frac{x}{y}}=\sqrt{\frac{xy}{y.y}}=\frac{\sqrt{xy}}{y}\)

d) \(\sqrt{\frac{9x^3}{25y}}=\sqrt{\frac{9x^3.y}{25y^2}}=\frac{3x\sqrt{xy}}{5y}\)

a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)

b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)

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

d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)

e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)

f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)

a) \(\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}.\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\)

b) \(\sqrt{\frac{3a}{5b}}=\frac{\sqrt{3a}}{\sqrt{5b}}=\frac{\sqrt{3a}.\sqrt{5b}}{5b}=\frac{\sqrt{15ab}}{5b}\left(a;b>0\right)\)

c) \(\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{12}}=\frac{\sqrt{5}.\sqrt{12}}{12}=\frac{\sqrt{60}}{12}=\frac{2\sqrt{15}}{12}=\frac{\sqrt{15}}{6}\)

d) \(\sqrt{\frac{5x}{18y}}=\frac{\sqrt{5x}}{\sqrt{18y}}=\frac{\sqrt{5x}}{\sqrt{3^2.2y}}=\frac{\sqrt{5x}}{3\sqrt{2y}}\)

\(=\frac{\sqrt{5x}.\sqrt{3y}}{3.2y}=\frac{\sqrt{15xy}}{6xy}\)

Quên mất k ghi đk xy > 0

a) \(\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{2}\)                           b)\(\sqrt{\dfrac{3a}{5b}}=\dfrac{\sqrt{3a}.\sqrt{5b}}{\sqrt{5b}.\sqrt{5b}}=\dfrac{\sqrt{15ab}}{5b}\)       \(\sqrt{\dfrac{5}{12}}=\dfrac{\sqrt{5}.\sqrt{12}}{\sqrt{12}.\sqrt{12}}=\dfrac{\sqrt{60}}{12}\)                        d)             

Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)

         \(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)                 (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)

         \(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

           \(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

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

Vậy giá trị biểu thức \(A=1.\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

bài này dài lắm mk ko tiện làm

mình cần gấp, thanks các bạn

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