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1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\)
\(\Rightarrow A=\dfrac{1}{1-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) ( Lượt \(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\) ở tử và mẫu )
\(\Rightarrow A=\dfrac{1}{1-\dfrac{1}{24}}\)
\(\Rightarrow A=\dfrac{1}{\dfrac{23}{24}}=\dfrac{24}{23}\)
Vậy \(A=\dfrac{24}{23}\)
\(\left(x^2-y^2\right)^2=\left(x-y\right)^2\left(x+y\right)^2\) \(\Rightarrow\left\{{}\begin{matrix}x;y>0\\x+y< 1\end{matrix}\right.\)=> dccm sai = > người ra đề sai họăc người chép đề sai ;
\(\cos a\times\sin b=-\dfrac{1}{2}\left[\sin\left(a-b\right)-\sin\left(a+b\right)\right]\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{-1}{2}\times1=-\dfrac{1}{2}\)
Quy đồng biểu thức ta được
=\(\dfrac{9+3\sqrt{6}-2\sqrt{6}-4+3\sqrt{6}-6+6+2\sqrt{6}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{5-2\sqrt{6}}{4}\)
=\(\dfrac{5+2\sqrt{6}}{1}\).\(\dfrac{5-2\sqrt{6}}{4}\)
=\(\dfrac{1}{4}\)
Lời giải:
$-1=\cos (a-b)=\cos a\cos b+\sin a\sin b$
$\Rightarrow -2=2\cos a\cos b+2\sin a\sin b$
Mà: $2=\cos ^2a+\sin ^2a+\cos ^2b+\sin ^2b$
Cộng theo vế 2 đẳng thức trên lại suy ra:
$0=(\cos a+\cos b)^2+(\sin a+\sin b)^2$
$\Rightarrow \cos a=-\cos b; \sin a=-\sin b$
$\frac{1}{2}=\sin (a+b)=\sin a\cos b-\cos a\sin b$
$=(-\sin b)(-\cos a)-\cos a\sin b=0$ (vô lý)
DO đó không tính được $\cos a\cos b$
Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\) \(\left(k\ne0\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=a.k\\y=b.k\\z=c.k\end{matrix}\right.\)
Ta có :
\(A=\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
\(A=\dfrac{\left[\left(a.k\right)^2+\left(b.k\right)^2+\left(c.k\right)^2\right]\cdot\left(a^2+b^2+c^2\right)}{\left(a.a.k+b.b.k+c.c.k\right)^2}\)
\(A=\dfrac{\left(a^2k^2+b^2k^2+c^2k^2\right)\cdot\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(A=1\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)