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a: \(E=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4\sqrt{x}\left(x-1\right)}{x-1}:\dfrac{x-1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4x\sqrt{x}-4\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
b: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
Thay x=2 vào E, ta được:
\(E=\dfrac{4\cdot2^2}{\left(2-1\right)^2}=16\)
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
Câu (A) đề có sao không nhỉ?
\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow\dfrac{1}{a-1}\)
\(E=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)
\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
ĐK : a >= 0 , a khác 1
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
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: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
E = ( \(\dfrac{\sqrt{x}}{\sqrt{x-1}}\)- \(\dfrac{1}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x+1}}\)+\(\dfrac{2}{\sqrt{x-1}}\))
a) ta có ĐKXĐ của E là x \(\ne\) 1
x \(\ne\) 0
x \(\ne\) -1
b) ( \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{\sqrt{1}}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x}+1}\)+\(\dfrac{2}{\sqrt{x}-1}\))
= (\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) :(\(\dfrac{1}{\sqrt{x}+1}\)+ \(\dfrac{2}{x+1}\))
= ( \(\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : (\(\dfrac{1\left(x-1\right)+2\sqrt{x+1}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\))
= ( \(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : \(\dfrac{x-1+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\): \(\dfrac{\left(x-1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\). \(\dfrac{1}{2}\)
= \(\dfrac{1}{2\sqrt{x}}\)
Lời giải:
ĐK: \(x\geq 0; x\neq 4;x\neq 9\)
a) Ta có:
\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+2+(x-9)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-3)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{x-4}\)
b)
Ta có: \(\frac{1}{P}\leq \frac{-5}{2}\)\(\Leftrightarrow \frac{x-4}{\sqrt{x}+1}\leq \frac{-5}{2}\)
\(\Leftrightarrow 2(x-4)\leq -5(\sqrt{x}+1)\)
\(\Leftrightarrow 2x+5\sqrt{x}-3\leq 0\)
\(\Leftrightarrow (2\sqrt{x}-1)(\sqrt{x}+3)\leq 0\)
\(\Rightarrow 2\sqrt{x}-1\leq 0\) (do \(\sqrt{x}+3>0\) )
\(\rightarrow x\leq \frac{1}{4}\)
Vậy \(0\leq x\leq \frac{1}{4}\)
a: ĐKXĐ: x>0
\(E=\dfrac{\sqrt{x}}{x+2\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+2+x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
b: E=2/5
=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)
=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)
=>\(2x-3\sqrt{x}+4=0\)
=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)
=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)
=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)
Vậy: \(x\in\varnothing\)