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\(a,\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=-1\)
\(b,\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(dk:x\ne\pm\sqrt{2}\right)\\ =\dfrac{x^2+2\sqrt{2}x+\sqrt{2^2}}{x^2-\sqrt{2^2}}\\ =\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\\ =\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\)
\(c,\sqrt{9x^2}-2x\left(dk:x< 0\right)\\ =\sqrt{3^2}.\sqrt{x^2}-2x\\ =3\left|x\right|-2x\\ =-3x-2x\\ =-5x\)
\(d,\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}-3+\sqrt{2}\\ =\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}\\ =\sqrt{2}+3-3+\sqrt{2}\\ =2\sqrt{2}\)
\(e,\dfrac{x^2-5}{x+\sqrt{5}}\left(dk:x\ne-\sqrt{5}\right)\\ =\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\\ =x-\sqrt{5}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{2}=\dfrac{4}{2}=2\)
b: \(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{15}}=\sqrt{\dfrac{6}{15}}=\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}\)
c: \(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\)
\(=3\sqrt{a}+9\sqrt{a}+3\cdot5\sqrt{a}-16\cdot7\sqrt{a}\)
\(=27\sqrt{a}-112\sqrt{a}=-85\sqrt{a}\)
d: \(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\sqrt{ab}-\sqrt{bc}}\)
\(=\sqrt{ab}+\sqrt{bc}\)
e: \(a\left(\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\cdot\sqrt{\dfrac{a}{b}}}\right)\cdot\sqrt{ab}\)
\(=a\cdot\sqrt{\dfrac{a}{b}\cdot ab+2\sqrt{ab}\cdot ab+b\cdot\sqrt{\dfrac{a}{b}}\cdot ab}\)
\(=a\cdot\sqrt{a^2+2\cdot ab\cdot\sqrt{ab}+a\sqrt{a}\cdot b\sqrt{b}}\)
\(=a\cdot\sqrt{a^2+3\cdot a\cdot\sqrt{a}\cdot b\cdot\sqrt{b}}\)
e: ĐKXĐ: a>=0 và a<>1
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\left(a-\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)\)
a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)
c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
Rút gọn
a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)
\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)
\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)
\(=\sqrt{2}\left(1+1\right)=2\sqrt{2}\)
b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)
\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)
\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)
\(=4\sqrt{a}-5\sqrt{10a}\)
c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-\sqrt{60}\)
\(=6-\sqrt{15}\)
a) \(ĐKXĐ:x\ne-\sqrt{3}\)
\(=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
b) \(=\dfrac{1-\sqrt{a^3}}{1-\sqrt{a}}=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
ĐKXĐ \(x^2+5x+2\ge0\)
Có : \(\left(x^2+5x+4\right)-3.\sqrt{x^2+5x+2}=6\)
<=> \(\left(x^2+5x+2\right)-3\sqrt{x^2+5x+2}-4=0\)
<=> \(\left(\sqrt{x^2+5x+2}-4\right)\left(\sqrt{x^2+5x+2}+1\right)=0\)
<=> \(\sqrt{x^2+5x+2}-4=0\) (vì \(\sqrt{x^2+5x+2}+1>0\forall x\))
<=> x2 + 5x - 14 = 0
<=> (x + 7)(x - 2) = 0
<=> \(\left[{}\begin{matrix}x=-7\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)