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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 \(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{\left(\sqrt{a}+\sqrt{3}\right)\left(\sqrt{a}-\sqrt{3}\right)}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)
b)\(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{ab}{a-b}=ab\)
a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)
Xét: \(\dfrac{2\left(a+b\right)}{\sqrt{a^3}-2\sqrt{2b^3}}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{2\left(a+b\right)}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{a+2b+\sqrt{2ab}}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}=\dfrac{1}{\sqrt{a}-\sqrt{2b}}\)\(\dfrac{\sqrt{a^3}+2\sqrt{2b^3}}{2b+\sqrt{2ab}}-\sqrt{a}=\dfrac{\left(\sqrt{a}+\sqrt{2b}\right)\left(a-\sqrt{2ab}+2b\right)}{\sqrt{2b}\left(\sqrt{a}+\sqrt{2b}\right)}-\sqrt{a}=\dfrac{\left(\sqrt{a}-\sqrt{2b}\right)^2}{\sqrt{2b}}\)\(\Rightarrow P=\dfrac{\sqrt{a}-\sqrt{2b}}{\sqrt{2b}}=\sqrt{\dfrac{a}{2b}}-1\)
b, Tự lm nhé.
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
a) Sai đề.
\(\dfrac{a+b}{b^2}\sqrt[]{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\left|a\right|}{\left|a+b\right|}=\left|a\right|\)
b) Sai đề.
\(\dfrac{a\sqrt[]{b}+b\sqrt[]{a}}{\sqrt[]{ab}}:\dfrac{1}{\sqrt[]{a}-\sqrt[]{b}}=\dfrac{\sqrt[]{ab}\left(\sqrt[]{a}+\sqrt[]{b}\right)}{\sqrt[]{ab}}.\left(\sqrt[]{a}-\sqrt[]{b}\right)=a-b\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
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.
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
@Nhã Doanh
giúp mk vs