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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 : \(0< a< 1\)
a: ĐKXĐ: a>=0; a<>1
b: \(A=\left(\dfrac{a+3\sqrt{a}+2}{3\sqrt{a}-2}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\left(a-1\right)\left(\sqrt{a}+2\right)-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{a-1}{2\sqrt{a}}\)
\(=\dfrac{a\sqrt{a}+2a-\sqrt{a}-2-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(=\dfrac{\left(a\sqrt{a}-a+\sqrt{a}-2\right)}{3\sqrt{a}-2}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(B=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
\(A=\dfrac{a-\sqrt{a}-6}{4-a}-\dfrac{1}{\sqrt{a}-2}=\dfrac{a+2\sqrt{a}-3\sqrt{a}-6}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}-\dfrac{1}{\sqrt{a}-2}=\dfrac{\sqrt{a}-3}{2-\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}=\dfrac{\sqrt{a}-2}{2-\sqrt{a}}=-1\) \(F=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1-\sqrt{a}}{1-a}.\dfrac{1-\sqrt{a}}{1-a}=\left(a+2\sqrt{a}+1\right).\dfrac{\left(1-\sqrt{a}\right)^2}{\left(\sqrt{a}+1\right)^2\left(1-\sqrt{a}\right)^2}=1\)
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
P/s gọi a = x cho dễ viết nhé
a, Với \(x\ge0;x\ne1;x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
chỗ này mình nghĩ ko phải trục căn thức đâu ha :D
b, Ta có P > 1/6 hay \(\frac{\sqrt{x}-2}{3\sqrt{x}}>\frac{1}{6}\Leftrightarrow\frac{\sqrt[]{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{6\sqrt{x}-12-3\sqrt{x}}{18\sqrt{x}}>0\Leftrightarrow\frac{3\sqrt{x}-12}{18\sqrt{x}}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)( vì \(18\sqrt{x}>0\))
\(\Leftrightarrow3\sqrt{x}>12\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Vậy \(x>16\)
cho mình hỏi đề có sai ko ? \(P< \frac{1}{6}\)mình nghĩ sẽ hợp lí hơn
んuﾘ ｲ hãy thuận theo ý thầy :)) và nhớ chú ý đến ĐKXĐ
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\left(\frac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để P > 1/6 thì \(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
<=> \(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
<=> \(\frac{2\sqrt{a}-4}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\)
<=> \(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Dễ thấy \(6\sqrt{a}>0\forall x>0\)
=> \(\sqrt{a}-4>0\)<=> \(\sqrt{a}>4\)<=> \(a>16\)
Vậy với a > 16 thì P > 1/6
Bài 2:
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(P=\left[\dfrac{\left(a-1\right)^2}{4a}\right].\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\sqrt{a}-1}{a-1}\right)\)
\(P=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{2\sqrt{a}.\left(-2\right)}{a-1}\)
\(P=\dfrac{\left(a-1\right)^2\left(-4\sqrt{a}\right)}{4a.\left(a-1\right)}\)
\(P=\dfrac{\left(a-1\right).\left(-\sqrt{a}\right)}{a}=\dfrac{-a\sqrt{a}+\sqrt{a}}{a}\)
Bài 1:
\(A=\dfrac{2}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)\(A=\dfrac{2\sqrt{2}}{2}-\dfrac{1\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1}\)
\(A=\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{1}+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(A=\sqrt{2}-\sqrt{3}-\sqrt{2}+\sqrt{3}+1\)
\(A=1\)
a. \(A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
\(A=\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(A=\dfrac{5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}}{5-4}-\dfrac{2+\sqrt{3}}{4-3}\)
\(A=\sqrt{3}-2-\sqrt{3}=-2\)
b.
\(B=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(B=\left[\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right].\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right).\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\dfrac{-8\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}\)
\(B=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
\(a.A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\sqrt{3}-\dfrac{1}{2-\sqrt{3}}=\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}\)
\(b.\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\left(a>0;a\ne4\right)\)
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
Lời giải:
ĐKXĐ: $a\geq 0; a\neq 4$
\(A=\left[\frac{\sqrt{a}(\sqrt{a}-2)-\sqrt{a}(\sqrt{a}+2)}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}\right].(\sqrt{a}+2)\)
\(=\frac{-4\sqrt{a}+4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{1}{2-\sqrt{a}}\)