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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
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
b)Áp dụng BĐT AM-GM ta có:
\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)
Xảy ra khi \(a=b\)
c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:
\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)
Xảy ra khi \(a=b\)
a)\(\dfrac{a^2+3}{\sqrt{a^2+3}}=\sqrt{a^2+3}\ge\sqrt{3}< 2\)\
sai đề
a: \(=4\left|a-3\right|=4\left(a-3\right)=4a-12\)
b: \(=9\cdot\left|a-9\right|=9\left(9-a\right)=81-9a\)
c: \(a^3b^6\cdot\sqrt{\dfrac{3}{a^6b^4}}=a^3b^6\cdot\dfrac{\sqrt{3}}{-a^3b^2}=-b^4\sqrt{3}\)
d: \(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{a-b}\)
\(=\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)