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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.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
a)
Ta có: \(9x=5y=15z\Rightarrow\dfrac{9x}{45}=\dfrac{5y}{45}=\dfrac{15z}{45}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{3}\Rightarrow\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}_{\left(1\right)}\)
và \(-x+y-z=11_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ só bằng nhau có:
\(\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}=\dfrac{-x+y-z}{-5+9-3}=\dfrac{11}{1}=11.\)
Từ đó: \(\left\{{}\begin{matrix}\dfrac{-x}{-5}=11\Rightarrow-x=-55\Rightarrow x=55.\\\dfrac{y}{9}=11\Rightarrow y=99.\\\dfrac{z}{3}=11\Rightarrow z=33.\end{matrix}\right.\)
Vậy.....
b); c); d); e) làm tương tự.
a: \(2^{300}=8^{100}< 9^{100}=3^{200}\)
b: Để E là số nguyên thì a-2+3 chia hết cho a-2
=>\(a-2\in\left\{1;-1;3;-3\right\}\)
hay \(a\in\left\{3;1;5;-1\right\}\)
d: =>3x-5=0 và 3y+0,4=0
=>x=5/3 và y=-0,4/3=-2/15
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Bài 1:
a: \(M=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)
b: \(N=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)
\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)
\(=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
Bài 1:
\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=2007.\dfrac{1}{90}-3\)
\(=19,3\)
Vậy S = 19,3
5b)\(S=1+3+3^2+...+3^{2013}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{2014}\)
\(\Rightarrow3S-S=3^{2014}-1\)
\(\Rightarrow S=\dfrac{3^{2014}-1}{2}\)
Bài 3:
Vì x,y,z tỉ lệ với 2;3;4 nên x/2=y/3=z/4
Đặt x/2=y/3=z/4=k
=>x=2k; y=3k; z=4k
\(M=\dfrac{5x+2y+z}{x+4y-3z}=\dfrac{10k+6k+4k}{2k+12k-12k}=10\)