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a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b) đề sai rồi nha
c) \(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}=\dfrac{a\sqrt{a}-4\sqrt{a}+2a-8}{a-4}\)
\(=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)\left(a-4\right)}{a-4}=\sqrt{a}+2\)
a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)
b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)
c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)
d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)
a: \(=\dfrac{-a-2\sqrt{a}+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(=\dfrac{-4a-6\sqrt{a}+4}{a-4}\cdot\dfrac{-\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+2}\)
\(=\dfrac{4a+6\sqrt{a}-4}{\sqrt{a}+2}\cdot\dfrac{\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(4a+6\sqrt{a}-4\right)}{4-a}\)
b: Để \(A=\sqrt{a}+2\) thì \(4a\sqrt{a}+6a-4\sqrt{a}=\left(\sqrt{a}+2\right)\left(4-a\right)=4\sqrt{a}-a\sqrt{a}+8-2a\)
=>\(5a\sqrt{a}+8a-8\sqrt{a}-8=0\)
=>\(5a\cdot\sqrt{a}+10a-2a-4\sqrt{a}-4\sqrt{a}-8=0\)
=>\(\left(\sqrt{a}+2\right)\left(5a-2\sqrt{a}-4\right)=0\)
=>\(5a-2\sqrt{a}-4=0\)
=>\(a=\dfrac{22+2\sqrt{21}}{25}\)
a) \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\)
= \(\sqrt{\dfrac{6+2\sqrt{5}}{4x^2}}-\sqrt{\dfrac{6-2\sqrt{5}}{4}}=\sqrt{\dfrac{5+2\sqrt{5}+1}{4x^2}}-\sqrt{\dfrac{5-2\sqrt{5}+1}{4}}\) = \(\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{\left(2x\right)^2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2^2}}=\dfrac{\left|\sqrt{5}+1\right|}{\left|2x\right|}-\dfrac{\left|\sqrt{5}-1\right|}{2}=\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\)
Thay x = 1 vào biểu thức \(\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\) ta được :
\(\dfrac{\sqrt{5}+1}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=1\)
Vậy tại x =1 thì giá trị của biểu thức \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\) là bằng 1
b) \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\)
= \(\sqrt{\dfrac{a\left(a^2+4a+4\right)}{a\left(a^2-2ab+b^2\right)}}-\sqrt{\dfrac{b\left(b^2-4b+4\right)}{b\left(a^2-2ab+b^2\right)}}+ab\)
= \(\dfrac{\sqrt{\left(a+2\right)^2}}{\sqrt{\left(a-b\right)^2}}-\dfrac{\sqrt{\left(b-2\right)^2}}{\sqrt{\left(a-b\right)^2}}+ab=\dfrac{a+2}{a-b}-\dfrac{b-2}{a-b}+ab\) = a - b + ab
Thay a = 4 và b = 3 vào biểu thức a - b +ab ta được :
4 - 3 + 4.3 = 13
Vậy tại a = 4 ; b = 3 thì giá trị của biểu thức \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\) là bằng 13
c) \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab=ab^2.\dfrac{2}{ab^2}+ab=2+ab\)
Thay a = 1 và b = -2 vào BT : 2 + ab ta được :
2 + 1.(-2) = 2 + (-2) = 0
Vậy tại a = 1 ; b = -2 thì giá trị của biểu thức \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab\) là bằng 0
d) \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) = \(\dfrac{a+b}{b^2}.\dfrac{\sqrt{a^2b^2}}{\sqrt{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{ab}{a+b}=\dfrac{ab}{b^2}\)
Thay a = 1 ; b =2 vào BT : \(\dfrac{ab}{b^2}\) ta được : \(\dfrac{1.2}{2^2}=\dfrac{1}{2}\)
Vậy tại a =1 ; b =2 GT của BT : \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) là \(\dfrac{1}{2}\)
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) \(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{4a+2\sqrt{a}-4}{4-a}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{2\sqrt{a}-a}\right)=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\left[\dfrac{-2\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right]=\left[\dfrac{a+4\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{2\sqrt{a}-a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{a+4\sqrt{a}+4+2\sqrt{a}-a-4a-2\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{2-\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{-4a+4\sqrt{a}+8}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{1}{\sqrt{a}}=\dfrac{4\left(2-\sqrt{a}\right)\left(\sqrt{a}+1\right).\sqrt{a}}{\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)}=\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}\)
Ta có A=\(\sqrt{a}+2\Leftrightarrow\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}=\sqrt{a}+2\Leftrightarrow4a+4\sqrt{a}=\left(\sqrt{a}+2\right)^2\Leftrightarrow4a+4\sqrt{a}=a+4\sqrt{a}+4\Leftrightarrow3a=4\Leftrightarrow a=\dfrac{4}{3}\left(tm\right)\)Vậy a=\(\dfrac{4}{3}\) thì A=\(\sqrt{a}+2\)
\(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{2+\sqrt{a}}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)\(=\left(\dfrac{\left(2+\sqrt{a}\right)^2+\sqrt{a}\left(2-\sqrt{a}\right)-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right)\)\(:\left(\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(=\dfrac{4+4\sqrt{a}+a+2\sqrt{a}-a-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) . \(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\dfrac{-4a+8\sqrt{a}}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) .\(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}.\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4a}{2+\sqrt{a}}\)
b, Để A=\(\sqrt{a}+2\)
<=> \(\dfrac{4a}{2+\sqrt{a}}\) =\(\sqrt{a}+2\)
<=> 4a=\(\left(\sqrt{a}+2\right)^2\)
<=> \(a+4\sqrt{a}+4-4a=0\)
<=> \(-3a+4\sqrt{a}+4=0\)
<=>\(-3a+6\sqrt{a}-2\sqrt{a}+4=0\)
<=> \(-3\sqrt{a}\left(\sqrt{a}-2\right)-2\left(\sqrt{a}-2\right)=0\)
<=> \(\left(\sqrt{a}-2\right)\left(-3\sqrt{a}-2\right)=0\)
<=>\(\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{-2}{3}\left(vl\right)\end{matrix}\right.\)
<=> a=4
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: \(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{a^2}\)
2: \(\dfrac{4a-1}{a^2}=\dfrac{4a-16+15}{a^2}=4\cdot\dfrac{\left(a-4\right)}{a^2}+\dfrac{15}{a^2}\)