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Với mọi \(k\ge2\) thì \(\frac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}=\frac{\left[\left(\sqrt{k-1}\right)^2+\left(\sqrt{k+1}\right)^2+\sqrt{\left(k-1\right)\left(k+1\right)}\right]\left(\sqrt{k+1}-\sqrt{k-1}\right)}{\left(\sqrt{k-1}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k-1}\right)}\)
\(=\frac{\sqrt{\left(k+1\right)^3}-\sqrt{\left(k-1\right)^3}}{2}\)
Suy ra tổng đã cho có thể viết là :
\(A=\frac{1}{2}\left[\sqrt{3^3}-\sqrt{1^3}+\sqrt{4^3}-\sqrt{2^3}+\sqrt{5^3}-\sqrt{3^3}+\sqrt{6^3}-\sqrt{4^3}+...+\sqrt{101^3}-\sqrt{99^3}\right]\)
\(=\frac{1}{2}\left[-1-\sqrt{2^3}+\sqrt{101^3}+\sqrt{100^3}\right]\)
\(=\frac{999+\sqrt{101^3}-\sqrt{8}}{2}\)
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)
b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)
\(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)
a)
\(A=2^{2-3\sqrt{5}}.8^{\sqrt{5}}=2^{2-3\sqrt{5}}.2^{3\sqrt{5}}=2^{\left(2-3\sqrt{5}\right)+3\sqrt{5}}=2^2=4\)
\(A=4\)
d)
\(D=\left(4^{2\sqrt{3}}-4^{\sqrt{3}-1}\right).2^{-2\sqrt{3}}=2^{4\sqrt{3}-2\sqrt{3}}-2^{2\sqrt{3}-2-2\sqrt{3}}\)
\(D=2^{2\sqrt{3}}-\dfrac{1}{4}\)
b) \(=\dfrac{3^{1+2\sqrt[3]{2}}}{3^{2\sqrt[3]{2}}}=3^{1+2\sqrt[3]{2}-2\sqrt[3]{2}}=3^1=3\)
c) \(=\dfrac{\left(2.5\right)^{2+\sqrt{7}}}{2^{2+\sqrt{7}}5^{1+\sqrt{7}}}=\dfrac{2^{2+\sqrt{7}}5^{2+\sqrt{7}}}{2^{2+\sqrt{7}}5^{1+\sqrt{7}}}=5\)
d) \(=\left(2^{2.\left(2\sqrt{3}\right)}-2^{2\left(\sqrt{3}-1\right)}\right).2^{-2\sqrt{3}}\)
\(=2^{4\sqrt{3}-2\sqrt{3}}-2^{2\sqrt{3}-2-2\sqrt{3}}\)
\(=2^{2\sqrt{3}}-2^{-2}\)
\(=2^{2\sqrt{3}}-\dfrac{1}{2^2}\)
\(=\dfrac{2^{2+2\sqrt{3}}-1}{4}\)
\(A=\log_a\left(a^2\sqrt[4]{a^3\sqrt[5]{a}}\right)=\log_a\left(a^2\sqrt[4]{a^3.a^{\frac{1}{5}}}\right)=\log_a\left[a^2\left(a^{\frac{16}{5}}\right)^{\frac{1}{4}}\right]=\log_a\left(a^2.a^{\frac{4}{5}}\right)=\frac{14}{5}\)
Ta có: \(A=\frac{1}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{2x-\sqrt{x}+2}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-x-2\sqrt{x}+2x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+2}\)
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) (ĐK: \(x\ge0;x\ne\dfrac{1}{9}\))
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}\right)^2-1^2}\right]:\left[\dfrac{\left(3\sqrt{x}+1\right)\cdot1}{3\sqrt{x}+1}-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right]\)
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}\)
\(A=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3x+3\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(A=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)
b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)
Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)
\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)
\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)
c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)
d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)
e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
Đặt \(A=\sqrt[4]{2+\sqrt{5}+2\sqrt{2+\sqrt{5}}};B=\sqrt[4]{2+\sqrt{5}-2\sqrt{2+\sqrt{5}}}\)
\(\Rightarrow N=A+B\)
Ta có \(AB=\sqrt[4]{\left(2+\sqrt{5}\right)^2-4\left(\sqrt{2+\sqrt{5}}\right)}=1\)
và \(A^4+B^4=4+2\sqrt{5}\)
Suy ra \(A^4+B^4=2A^2B^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A^2+B^2=\sqrt{5}+1\)
Tức là :
\(A^2+B^2+2AB=\sqrt{5}+3=\left(\frac{\sqrt{5}+1}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow A+B=\frac{\sqrt{5}+1}{\sqrt{2}}\)
Vậy \(N=\frac{\sqrt{5}+1}{\sqrt{2}}\)