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) \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\)

= \(\sqrt{\dfrac{6+2\sqrt{5}}{4x^2}}-\sqrt{\dfrac{6-2\sqrt{5}}{4}}=\sqrt{\dfrac{5+2\sqrt{5}+1}{4x^2}}-\sqrt{\dfrac{5-2\sqrt{5}+1}{4}}\) = \(\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{\left(2x\right)^2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2^2}}=\dfrac{\left|\sqrt{5}+1\right|}{\left|2x\right|}-\dfrac{\left|\sqrt{5}-1\right|}{2}=\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\)

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 \(\dfrac{1}{2}\)

2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)

\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)

\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(a\ge0\)

\(\Rightarrow a\in\left\{0;2;3\right\}\)

Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)

\(=\dfrac{2}{a-1}\)

Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)

hay \(a\in\left\{2;0;3\right\}\)

a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\sqrt{3}+2+\sqrt{2}+1-\sqrt{2}-\sqrt{3}\)

=3

b) Ta có: \(B=\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left[\sqrt{3}+1-3\left(2+\sqrt{3}\right)+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right]\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{5}{2}\left(3+\sqrt{3}\right)\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(-5-2\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}=\dfrac{1}{2}\)

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}

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.