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5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
\(\dfrac{x}{z}=\dfrac{z}{y}\Rightarrow\dfrac{x.z}{z.y}=\dfrac{x}{y}=\dfrac{x^2}{z^2}=\dfrac{z^2}{y^2}=\dfrac{x^2+z^2}{y^2+z^2}\)
đăt \(\dfrac{x}{z}=\dfrac{z}{y}=k\)
=>\(\left\{{}\begin{matrix}x=zk\\z=yk\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=yk^2\\z=yk\end{matrix}\right.\)
ta có :\(\dfrac{x}{y}=\dfrac{yk^2}{y}=k^2\left(1\right)\)
lại có \(\dfrac{x^2+z^2}{y^2+z^2}=\dfrac{y^2k^4+y^2k^2}{y^2+y^2k^2}=\dfrac{y^2k^2.\left(k^2+1\right)}{y^2.\left(1+k^2\right)}=k^2\left(2\right)\)
từ (1) và (2) => ĐPCM
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
\(1.\)
\(a.\)
\(\dfrac{x}{-150}=-\dfrac{6}{x}\)
\(\Rightarrow x^2=\left(-6\right)\left(-150\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\pm30\)
\(2.\)
\(a.\) \(2x=3y;5y=7z\) và \(3x-7y+5z=30\)
Ta có : \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) \(\left(1\right)\)
\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) \(\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
\(\Rightarrow\dfrac{x}{21}=2\Rightarrow x=42\)
\(\dfrac{y}{14}=2\Rightarrow y=28\)
\(\dfrac{z}{10}=2\Rightarrow z=20\)
Vậy : ..................
\(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\Leftrightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x^2}{18}=\dfrac{y^2}{16}=\dfrac{2x^2+y^2}{18+16}=\dfrac{136}{34}=4\)
Suy ra: \(\left\{{}\begin{matrix}x^2=4.9=36\\y^2=4.16=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)
2) Ta có: \(2^{20}=\left(2^4\right)^5=16^5\)
Được biết số có tận cùng là \(6\) thì lũy thừa bao nhiêu cũng bằng \(6\)
Nên \(16^5=\overline{...6}\Leftrightarrow16^5-1=\overline{.....5}⋮5\)
Nên \(\dfrac{2^{20}-1}{5}\) là số nguyên
3)
Ta có:
\(A=100^2+200^2+...+1000^2\)
\(A=\left(1.100\right)^2+\left(2.100\right)^2+...+\left(10.100\right)^2\)
\(A=1^2.100^2+2^2.100^2+....+10^2.100^2\)
\(A=100^2\left(1^2+2^2+...+100^2\right)\)
\(A=10000.385=3850000\)