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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 \(\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\cdot\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}\right)=\dfrac{a+b}{2\sqrt{ab}}\)
\(2\sqrt{x^2}-1=2\cdot\dfrac{a+b}{2\sqrt{ab}}-1=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\)
\(x-\sqrt{x^2-1}=\dfrac{a+b}{2\sqrt{ab}}-\sqrt{\dfrac{a^2+2ab+b^2}{4ab}-1}\)
\(=\dfrac{a+b}{2\sqrt{ab}}-\dfrac{a-b}{2\sqrt{ab}}=\dfrac{2b}{2\sqrt{ab}}=\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(G=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}:\dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\dfrac{a+b-\sqrt{ab}}{b}\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
\(a.A=\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{x}\right)^2-\dfrac{2}{x}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(\dfrac{x+1}{x}\right)^2-2.\dfrac{x+1}{x}.\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{x}-\dfrac{1}{x+1}\right)^2}=\left|x+\dfrac{1}{x}+\dfrac{1}{x+1}\right|\)
\(b.\) Áp dụng điều đã CM ở câu a , ta có :
\(B=\sqrt{1+\dfrac{1}{1^1}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=100-\dfrac{1}{100}=\)
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a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)