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Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
Nếu là z+x thì mik biết làm nè:
Đặt x-y=2011(1)
y-z=-2012(2)
z+x=2013(3)
Cộng (1);(2);(3) lại với nhau ta được :
2x=2012=>x=1006
Từ (1) => y=-1005
Từ (3) => z=1007
B5
a)\(A=\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot\left(1-\dfrac{2010}{2010}\right)\left(1-\dfrac{2011}{2010}\right)\\ =\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot\left(1-1\right)\left(1-\dfrac{2011}{2010}\right)\\ =\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot0\cdot\left(1-\dfrac{2011}{2010}\right)\\ =0\)
b)
\(A=\dfrac{1946}{1986}=\dfrac{1986-40}{1986}=\dfrac{1986}{1986}-\dfrac{40}{1986}=1-\dfrac{40}{1986}\\ B=\dfrac{1968}{2008}=\dfrac{2008-40}{2008}=\dfrac{2008}{2008}-\dfrac{40}{2008}=1-\dfrac{40}{2008}\)
Vì \(\dfrac{40}{1986}>\dfrac{40}{2008}\) nên \(1-\dfrac{40}{1986}< 1-\dfrac{40}{2008}\) hay \(A< B\)
B6
a) Đề sai
Sửa lại:
\(B=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{28\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{28}-\dfrac{1}{31}\\ =1-\dfrac{1}{31}\\ =\dfrac{30}{31}\)
b)
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
Ta thấy:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=\dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}=\dfrac{1}{7}-\dfrac{1}{8}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\\ B< 1-\dfrac{1}{8}\\ B< \dfrac{7}{8}\left(1\right)\)
Mà \(\dfrac{7}{8}< 1\left(2\right)\)
Từ (1) và (2) ta có \(B< 1\)
a: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>x=12; y2=1; z3=-8
=>x=12; \(y\in\left\{1;-1\right\}\); z=-2
b: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{z}{-17}=\dfrac{t}{9}\)
=>x/5=y/-3=z/-17=t/9=-2
=>x=-10; y=6; z=34; t=-18
\(\left(3n\right)^{100}\\ =3^{100}.n^{100}\\ =\left(3^4\right)^{25}.n^{100}\\ =81^{25}.n^{100}⋮81\)
Vậy \(\left(3n\right)^{100}⋮81\)
Chúc em học tốt!
Tuy có vẻ hơi muộn nhưng thôi
Nếu A là số tự nhiên ⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
\(\Rightarrow7^{2004}-3^{92^{94}}⋮10\)
Thật vậy, ta có :
72004 với lũy thừa là 2004 ⋮ 4
⇒ 72004 = ( .......... 9 )
392^94 với lũy thừa là 9294 mà 92 ⋮ 4 ⇒ 9294 ⋮ 4
⇒ 392^94 = ( .......... 9 )
⇒ 72004 - 392^94 = ( .......... 9 ) - ( ............ 9) = ( ........... 0 ) ⋮ 10
⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
A=1/10.(72004-392^94) là số tự nhiên.
Bài 8:
a: \(=\dfrac{5}{7}-\dfrac{3}{5}=\dfrac{25-21}{35}=\dfrac{4}{35}\)
b: \(=\dfrac{-3}{7}+\dfrac{2}{9}=\dfrac{-27+14}{63}=\dfrac{-13}{63}\)
c: \(=\dfrac{-5}{11}+\dfrac{6}{7}=\dfrac{-35+66}{77}=\dfrac{31}{77}\)
d: \(=\dfrac{334}{335}-\dfrac{4}{335}=\dfrac{330}{335}=\dfrac{66}{67}\)