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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) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Bài 1:
a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
b: \(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)
=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)
=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)
=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)
b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)
\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)
e: \(=2-x-x=2-2x\)
i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)
\(=\left|2x-1\right|-2x\)
=1-2x-2x=1-4x
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)