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Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+......................+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.....................+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...................+\dfrac{101}{3^{101}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..............+\dfrac{100}{3^{99}}\right)\)\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+................+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+.............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)-\left(1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Chúc bn học tốt ~
a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)
= \(\dfrac{-7}{9}.\dfrac{11}{4}\)
= \(\dfrac{-77}{36}\)
b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)
= \(\dfrac{2}{3}+\dfrac{-2}{15}\)
= \(\dfrac{10}{15}+\dfrac{-2}{15}\)
= \(\dfrac{-8}{15}\)
c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)
= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)
= \(\dfrac{2}{3}-5\)
= \(\dfrac{-13}{3}\)
d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)
= \(\dfrac{1}{2}\) . 11 - 7
= \(\dfrac{11}{2}-\dfrac{14}{2}\)
= \(\dfrac{-3}{2}\)
e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)
= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)
= \(\dfrac{3}{4}.-28\)
= \(-21\)
Bài 1
a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)
\(=-\dfrac{24}{33}\)
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
D = \(\dfrac{1}{2}+\dfrac{ 1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\)
D = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)
2x D = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\)
2D - D = 1 - \(\dfrac{1}{2^{100}}\)
D = 1 - \(\dfrac{1}{2^{100}}\)
\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\\ =>2D=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ =>2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ =>D=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)