\(\lim\limits\dfrac{\sqrt{2\cdot4^n+1}-2^n}{\sqrt{2\cdot4^n+1}+2^n}\)

\(=\lim\limits\dfrac{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}-2^n}{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}+2^n}\)

\(=\lim\limits\dfrac{\sqrt{2+\dfrac{1}{4^n}}-1}{\sqrt{2+\dfrac{1}{4^n}}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)

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

=>a=3; b=-2

\(a^3+b^3=3^3+\left(-2\right)^3=27-8=19\)

3

37:

\(\lim\limits_{x\rightarrow1}\dfrac{x^4-x^3-x+1}{x^3-5x^2+7x-3}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{x^3\left(x-1\right)-\left(x-1\right)}{x^3-x^2-4x^2+4x+3x-3}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^3-1\right)}{\left(x-1\right)\left(x^2-4x+3\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\cdot\left(x^2+x+1\right)}{\left(x-1\right)\left(x-3\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{x^2+x+1}{x-3}=\dfrac{1+1+1}{1-3}=\dfrac{3}{-2}=-\dfrac{3}{2}\)

a: Ta có: \(\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=90^0+\widehat{BAC}\)

\(\widehat{NAB}=\widehat{BAC}+\widehat{NAC}=\widehat{BAC}+90^0\)

Do đó: \(\widehat{MAC}=\widehat{NAB}\)

Xét ΔMAC và ΔBAN có

MA=BA

\(\widehat{MAC}=\widehat{BAN}\)

AC=AN

Do đó: ΔMAC=ΔBAN

b: Gọi H là giao điểm của CM và BN

Ta có: ΔMAC=ΔBAN

=>\(\widehat{ANB}=\widehat{ACM}\)

=>\(\widehat{ANH}=\widehat{ACH}\)

=>AHCM là tứ giác nội tiếp

=>\(\widehat{NHC}=\widehat{NAC}=90^0\)

=>NB\(\perp\)MC tại H

\(\lim\limits\dfrac{u_n+1}{3\cdot u_n^2+5}\)

\(=\lim\limits\dfrac{\dfrac{1}{u_n}+\dfrac{1}{u_n^2}}{3+\dfrac{5}{u_n^2}}\)

\(=\dfrac{0+0}{3+0}=\dfrac{0}{3}=0\)

\(\dfrac{u_2}{q\left(1-q\right)}=32\rightarrow\dfrac{8}{q\left(1-q\right)}=32\rightarrow q=\dfrac{1}{2}\rightarrow u_1=16\\ \rightarrow u_n=16\left(\dfrac{1}{2}\right)^{n-1}=2^{5-n}\)