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a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
\(a.\sqrt{\left(1-\sqrt{5}\right)^2}+1=\left|1-\sqrt{5}\right|+1=\sqrt{5}-1+1=\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}-2=\sqrt{\left(\sqrt{2}+1\right)^2}-2=\sqrt{2}+1-2=\sqrt{2}-1\)
\(c.\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)=\sqrt{\left(b-\dfrac{1}{2}\right)^2}-2b+\dfrac{1}{2}=b-\dfrac{1}{2}-2b+\dfrac{1}{2}=-2b\)
\(d.\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}\)
\(e.\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-2\)
\(g.3x+\sqrt{x^2-2x+1}=3x+\sqrt{\left(x-1\right)^2}\)
* \(x\ge1\Rightarrow3x+\left|x-1\right|=3x+x-1=4x-1\)
* \(x< 1\Rightarrow3x+\left|x-1\right|=3x+1-x=2x+1\)
\(h.\sqrt{y+2\sqrt{y^2-2y+1}}=\sqrt{y+2\sqrt{\left(y-1\right)^2}}=\sqrt{y+2y-2}=\sqrt{3y-2}\left(y\ge1\right)\) hoặc: \(\sqrt{y+2-2y}=\sqrt{-y+2}\left(y< 1\right)\)
\(H=\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
\(H^2=17-2\sqrt{32}+17+2\sqrt{32}+2\sqrt{\left(17-2\sqrt{32}\right)\left(17+2\sqrt{32}\right)}=34+2\sqrt{161}\)
\(H=\sqrt{34+2\sqrt{161}}\)
\(k.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
1. \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\sqrt{a}-2\)
= 0
2: \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)
4: \(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\dfrac{1}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}=\sqrt{a}+1\)
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Biết đâu làm đó , sai thôi đừngg chửi nhé
1, Rút gọn
a) A = \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\) = \(\dfrac{\left(\sqrt{x}\right)^2+\sqrt{xy}}{\left(\sqrt{y}\right)^2+\sqrt{xy}}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}\) = \(\dfrac{\sqrt{x}}{\sqrt{y}}\)
b) B = \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) . \(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\)
= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}.\dfrac{bc^3}{\left(a-b\right)}}\) = \(\sqrt{\left(a-b\right)^2.b^4.c^2}\)
= \(\left|a-b\right|\) . \(\left|b^2\right|\) . \(\left|c\right|\)
= -(a -b) .b2. c
bài 2:
a/ \(\sqrt{x^2-4}-\sqrt{x-2}=0\) đk: x≥2
<=> \(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
<=>\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
vậy pt có 1 nghiệm x = 2
b/ \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)
Ta có: \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=\sqrt{3\left(x^2+4x+4\right)+4}+\sqrt{\left(y^2-4y+4\right)+9}=\sqrt{3\left(x+2\right)^2+4}+\sqrt{\left(y-2\right)^2+9}\ge\sqrt{4}+\sqrt{9}=2+3=5\)
=> Dấu ''='' xảy ra khi x = -2; y = 2
Vậy pt có nghiệm x=-2; y = 2
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)