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\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
\(=x-3\sqrt{x}-5\sqrt{x}-15\)
\(=x-8\sqrt{x}-15\)
b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)
\(=-x+2\sqrt{x}+12\)
c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
\(=x-9-5\sqrt{x}+5\)
\(=x-5\sqrt{x}-4\)
d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3\sqrt{x}-6-x+1\)
\(=-x+3\sqrt{x}-5\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)
b. \(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=x-y\)
c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)
\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{3}\right)^2-5=1+2\sqrt{3}+3-5=2\sqrt{3}-1\)
em cảm mơn nhiều ạ