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16 tháng 5 2023
\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)
\(=\pm3\sqrt{2}x^2\)
\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)
\(=\pm\left|a\right|\)
17 tháng 11 2017
a, = \(\sqrt{a^2b^2.\left(1+\frac{1}{a^2b^2}\right)}\) = \(\sqrt{a^2b^2+1}\)
c, = \(\sqrt{\frac{a+ab}{b^4}}\) = \(\frac{\sqrt{a+ab}}{b^2}\)
k mk nha
17 tháng 11 2017
a, \(ab\sqrt{1+\frac{1}{a^2b^2}}\)
\(ab\sqrt{1+\frac{1}{a^2b^2}}=ab\sqrt{\frac{1+a^2b^2}{a^2b^2}}=\frac{ab}{\left|ab\right|}\sqrt{1+a^2b^2}\)
\(=\hept{\begin{cases}\sqrt{1+a^2b^2}ĐK:ab>0\\-\sqrt{1+a^2b^2}ĐKab< 0\end{cases}}\)
b, \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}\)
\(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a+ab}{b^4}}=\frac{1}{b^2}\sqrt{a+ab}\)
19 tháng 8 2019
\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}.\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
19 tháng 8 2019
\(B=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\)\(\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
17 tháng 8 2016
a/ \(\sqrt{8\left(\sqrt{2}-\sqrt{3}\right)^2}=2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{6}-4\)
b/ \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab.\sqrt{\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2.\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2+1}\)
c/ \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a}{b^3}\left(1+\frac{1}{b}\right)}=\frac{1}{b}.\sqrt{\frac{a}{b}\left(1+\frac{1}{b}\right)}\)
d/ \(\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
A
20 tháng 6 2021
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\times\frac{\sqrt{x^2-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
Thay \(x=\frac{a^2+b^2}{2ab}\)vào A, ta được :
\(A=\frac{\sqrt{\frac{a^2+b^2}{2ab}+1}+\sqrt{\frac{a^2+b^2}{2ab}-1}}{\sqrt{\frac{a^2+b^2}{2ab}+1}-\sqrt{\frac{a^2+b^2}{2ab}-1}}\)
\(A=\frac{\sqrt{\frac{\left(a+b\right)^2}{2ab}}+\sqrt{\frac{\left(b-a\right)^2}{2ab}}}{\sqrt{\frac{\left(a+b\right)^2}{2ab}}-\sqrt{\frac{\left(b-a\right)^2}{2ab}}}\)
\(A=\frac{a+b\sqrt{\frac{1}{2ab}}+\left(b-a\right)\sqrt{\frac{1}{2ab}}}{a+b\sqrt{\frac{1}{2ab}}-\left(b-a\right)\sqrt{\frac{1}{2ab}}}\)
\(A=\frac{a+b+b-a}{a+b-b+a}\)
\(A=\frac{2b}{2a}\)
\(A=\frac{b}{a}\)
Ps : Nhớ k cho tui nhó, tui đã rất cố gắng rồi đấy. :)) K để lần sau có j tui giải giúp cho :)))
# Aeri #
23 tháng 7 2016
\(P=\left(\frac{a+\sqrt{a^2-b^2}}{a-\sqrt{a^2-b^2}}-\frac{a-\sqrt{a^2-b^2}}{a+\sqrt{a^2-b^2}}\right):\frac{4\sqrt{a^4-a^2b^2}}{b^2}\)
\(=\left[\frac{\left(a+\sqrt{a^2-b^2}\right)\left(a+\sqrt{a^2-b^2}\right)-\left(a-\sqrt{a^2-b^2}\right)\left(a-\sqrt{a^2-b^2}\right)}{\left(a-\sqrt{a^2-b^2}\right)\left(a+\sqrt{a^2-b^2}\right)}\right]:\frac{4\sqrt{a^2\left(a^2-b^2\right)}}{b^2}\)
\(=\left[\frac{\left(a+\sqrt{a^2-b^2}\right)^2-\left(a-\sqrt{a^2-b^2}\right)}{a^2-\left(a^2-b^2\right)}\right]:\frac{4a\sqrt{a^2-b^2}}{b^2}\)
\(=\frac{\left(a+\sqrt{a^2-b^2}+a-\sqrt{a^2-b^2}\right)\left(a+\sqrt{a^2-b^2}-a+\sqrt{a^2-b^2}\right)}{b^2}\cdot\frac{b^2}{4a\sqrt{a^2-b^2}}\)
\(=\frac{2a\cdot2\sqrt{a^2-b^2}}{b^2}\cdot\frac{b^2}{4a\sqrt{a^2-b^2}}\)
\(=1\)
Đặt A=\(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{\left(a-b\right)^2}}=\frac{a-b}{b^2}.\left|\frac{ab^2}{a-b}\right|\)
Với a<b thì : A=\(\frac{a-b}{b^2}.\frac{ab^2}{-\left(a-b\right)}=-a\)
Với a>b thì : A=\(\frac{a-b}{b^2}.\frac{ab^2}{a-b}=a\)
\(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)
\(=\frac{a-b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a-b\right)^2}}\)
\(=\frac{a-b}{b^2}\cdot\frac{\sqrt{\left(ab^2\right)^2}}{\sqrt{\left(a-b\right)^2}}\)
\(=\frac{a-b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a-b\right|}\)
+) Nếu a>b => \(\frac{a-b}{b^2}\cdot\frac{ab^2}{a-b}=a\)
+) Nếu a<b => \(\frac{a-b}{b^2}\cdot\frac{ab^2}{b-a}=-a\)