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đk : \(a\ge0;b\ge0;a\ne b\)
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{a+2\sqrt{ab}+b+a-2\sqrt{ab}+b}{a-b}\) = \(\dfrac{2\left(a+b\right)}{a-b}\)
b) đk : \(a\ge0;b\ge0;a\ne b\)
\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{1}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}\)
= \(\dfrac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)
a) ĐS: .
b) ĐS: Nếu thì
Nếu ab
c) ĐS:
d)
Nhận xét. Nhận thấy rằng để có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.
a) ĐS: .
b) ĐS: Nếu thì
Nếu ab
c) ĐS:
d)
Nhận xét. Nhận thấy rằng để có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.
Câu (A) đề có sao không nhỉ?
\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow\dfrac{1}{a-1}\)
\(E=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
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
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
a)
\(\sqrt{\dfrac{27a^4}{48a^2}}=\sqrt{\dfrac{9a^2}{16}}=\sqrt{\left(\dfrac{3a}{4}\right)^2}=\dfrac{3a}{4}\)
b)
\(\dfrac{\sqrt{9x^2-25}}{\sqrt{3x+5}}=\dfrac{\sqrt{\left(3x-5\right)\left(3x+5\right)}}{\sqrt{3x+5}}=\sqrt{3x-5}\)
c)
\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\\ =\left(3-\sqrt{5}\right)+2.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\\ =2.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+6\\ =2.\sqrt{9-5}+6\\ =10\\ \Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\)
d) KO khó!!
a) \(\sqrt{\dfrac{x}{y^3}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy}{y^4}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy+2x}{y^4}}=\dfrac{\sqrt{xy+2x}}{\sqrt{y^4}}=\dfrac{\sqrt{xy+2x}}{\left|y^2\right|}=\dfrac{\sqrt{xy+2x}}{y^2}\)(vì y2\(\ge0\))
b) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}.\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)
c) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}\)
Nếu a-b>0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{a-b}=\left|ab\right|\)
Nếu a-b<0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{-\left(a-b\right)}=-\left|ab\right|\)
d) \(\dfrac{a-3\sqrt{a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}\right)^3+\left(\sqrt{3}\right)^3}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-3\sqrt{a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)
Nếu trục căn thức ở mẫu thì \(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{\left(\sqrt{a}+\sqrt{3}\right)\left(\sqrt{a}-\sqrt{3}\right)}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)
\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)
\(=\sqrt{a}+\sqrt{b}-\left(a+\sqrt{ab}+b\right)\)
\(=\sqrt{a}+\sqrt{b}-a-\sqrt{ab}-b\)