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\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\)
Ta có: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
\(=a-b\)
Thay a = 2√3 và b = √3 vào P, ta được:
P = 2√3 - √3 = √3
Vậy...
a) Ta có: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\cdot\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}\cdot\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
\(=a-b\)
b) Thay \(a=2\sqrt{3}\) và \(b=\sqrt{3}\) vào biểu thức P=a-b, ta được:
\(P=2\sqrt{3}-\sqrt{3}=\sqrt{3}\)
Vậy: Khi \(a=2\sqrt{3}\) và \(b=\sqrt{3}\) thì \(P=\sqrt{3}\)
a: \(P=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{a-b}+b\right]\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)}+b\right]\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)}{a-\sqrt{ab}+b}\cdot\left(\dfrac{\sqrt{a}}{a+\sqrt{ab}+b}+b\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{\sqrt{a}+ab+b\sqrt{ab}+b^2}{a+\sqrt{ab}+b}\)
b: Thay a=16 và b=4 vào P, ta được:
\(P=\dfrac{4+2}{16-4\cdot2+4}\cdot\dfrac{4+16\cdot4+4\cdot4\cdot2+16}{16+4\cdot2+4}\)
\(=\dfrac{6}{12}\cdot\dfrac{116}{28}=\dfrac{29}{14}\)
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
\(P=\left(\dfrac{\sqrt{a}-b}{\sqrt{a}+b}-\dfrac{\sqrt{a}+b}{\sqrt{a}-b}\right)\cdot\left(\sqrt{a^3}-\dfrac{ab^2}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-b\right)^2-\left(\sqrt{a}+b\right)^2}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{\sqrt{a^4}-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-b-b-\sqrt{a}\right)\left(\sqrt{a}-b+b+\sqrt{a}\right)}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{a^2-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(-2b\right)\cdot\left(2\sqrt{a}\right)}{a-b^2}\cdot\dfrac{a\left(a-b^2\right)}{\sqrt{a}}\)
\(=\dfrac{-4b\sqrt{a}}{\sqrt{a}}\cdot a=-4ba\)
a: \(P=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)