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\(D=\log_{5^{-1}}\left(5^2\right)-3\log_{3^2}\left(3^{-1}\right)+4.\log_{2^{\frac{3}{2}}}2^6=-2+\frac{3}{2}+16=\frac{31}{2}\)
\(B=25^{\frac{1}{2}+\frac{1}{9}\log_{\frac{1}{2}}27+\log_{125}81}=\left(5^2\right)^{\frac{1}{2}+\frac{1}{9}\log_{5^{-1}}3^3+\log_{5^3}3^4}\)
\(=5^{1-\frac{2}{3}\log_53+\frac{8}{3}\log_53}=5^{1+2\log_53}=5.5^{\log_53^2}=5.9=45\)
Ta có : \(1+\left(\frac{x^4-1}{2x^2}\right)^2=\frac{x^8+2x^4+1}{4x^4}\) nên \(1+\sqrt{1+\left(\frac{x^4-1}{2x^2}\right)^2}=1+\frac{x^4+1}{2x^2}=\frac{\left(x^2+1\right)^2}{2x^2}\)
Do đó \(N=\frac{x^2+1}{x\sqrt{2}}\), thay \(x=\frac{1}{\sqrt{2}}\left(2^{\sqrt{2}}-2^{-\sqrt{2}}\right)\) vào ta được :
\(N=\frac{\frac{1}{2}\left(2^{\sqrt{2}}+2^{-\sqrt{2}}-2\right)+1}{\frac{1}{2}\left(2^{\sqrt{2}}+2^{-\sqrt{2}}\right)}=\frac{2^{2\sqrt{2}}+2^{-2\sqrt{2}}}{2^{\sqrt{2}}+2^{-\sqrt{2}}}\)
1. Xét x = - 2, thay vào pt ta dc: -1.0 = 4.0 (Hợp lí)
Vậy x = -2 là 1 nghiệm của pt
Xét x \(\ne\)- 2, ta có: x + 1 = 2 - x
<=> 2x = 1 <=> x = 1/2
Vậy S = {1/2; -2}
2. a. \(2\left(m+\frac{3}{5}\right)-\left(m+\frac{13}{5}\right)=5\)
<=> \(2m+\frac{6}{5}-m-\frac{13}{5}=5\)
<=> m = \(\frac{32}{5}\)
b. \(2\left(3m+1\right)+\frac{1}{4}-\frac{2\left(3m-1\right)}{5}+3m+\frac{1}{5}=5\)
<=> \(6m+2+\frac{1}{4}-\frac{6m-2}{5}+3m+\frac{1}{5}=5\)
<=> \(6m-\frac{6m-2}{5}+3m=5-2-\frac{1}{4}-\frac{1}{5}\)
<=> \(9m-\frac{6m-2}{5}=\frac{51}{20}\)
<=> \(\frac{45m-6m+2}{5}=\frac{51}{20}\)
<=> \(20\left(39m+2\right)=51.5\)
<=> 780m + 40 = 255
<=> 780m = 215
<=> m = \(\frac{43}{156}\)
a) \(A=\frac{a^{\frac{5}{2}}\left(a^{\frac{1}{2}}-a^{\frac{-3}{2}}\right)}{a^{\frac{1}{2}}\left(a^{\frac{-1}{2}}-a^{\frac{3}{2}}\right)}=\frac{a^3-a}{1-a^2}=-a\)
Do đó : \(A=-\left(\pi-3\sqrt{2}\right)=3\sqrt{2}-\pi\)
b) Rút gọn B ta có :
\(B=\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)\left[\left(a^{\frac{1}{3}}\right)^2+\left(b^{\frac{1}{3}}\right)^2\right]=\left(a^{\frac{1}{3}}\right)^3+\left(b^{\frac{1}{3}}\right)^3=a+b\)
Do đó :
\(B=\left(7-\sqrt{2}\right)+\left(\sqrt{2}+3\right)=10\)
\(B=\frac{a^{\frac{1}{4}}-a^{\frac{9}{4}}}{a^{\frac{1}{4}}-a^{\frac{5}{4}}}-\frac{b^{-\frac{1}{2}}-b^{\frac{3}{2}}}{b^{\frac{1}{2}}+b^{-\frac{1}{2}}}=\frac{a^{\frac{1}{4}}\left(1-a^2\right)}{a^{\frac{1}{4}}\left(1-a\right)}-\frac{b^{-\frac{1}{2}}\left(1-b^2\right)}{b^{-\frac{1}{2}}\left(1-b\right)}\)
\(=\left(1+a\right)-\left(1-b\right)=a+b=2013-\sqrt{2}+\sqrt{2}-2015=1\)
Ta có : \(C=\left(2^4.10^{-4}\right)^{-\frac{1}{4}}+3.64^{\frac{1}{12}}-\left(9-4\sqrt{2}\right)-7\sqrt{2}=5+3\sqrt{2}-9-3\sqrt{2}=-4\)
\(\frac{\left(\sqrt{5}-1\right)\left(6+2\sqrt{5}\right)}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)^2}{\sqrt{5}-1}=4\)
\(A=\log_3\left(\log_{2\sqrt{2}}\sqrt[3]{\sqrt{2}}\right)=\log_3\left(\log_{2^{\frac{3}{2}}}2^{\frac{1}{6}}\right)=\log_3\left(\frac{1}{6}.\frac{2}{3}\right)=\log_33^{-2}=-2\)
