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HL
5 tháng 7 2017
a) \(x-25\%x=0,5\)
\(\dfrac{3}{4}x=0,5\)
x = \(\dfrac{2}{3}\)
b) \(\left(50\%x+5\dfrac{1}{4}\right).\dfrac{-2}{3}=2\dfrac{5}{6}\)
\(\left(0,5x+\dfrac{21}{4}\right)=\dfrac{-17}{4}\)
\(0,5x=\dfrac{-19}{2}\)
x = -19
c) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)+2x=1,5:\dfrac{3}{5}\)
\(\dfrac{2}{3}+2x=\dfrac{5}{2}\)
\(2x=\dfrac{11}{6}\)
x= \(\dfrac{11}{12}\)
Q
14 tháng 4 2017
Bài 1: Tìm x biết:
a) \(\dfrac{6}{5}-2\left|1-3x\right|=1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{6}{5}-1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{-7}{15}\)
\(\left|1-3x\right|=\dfrac{-7}{15}:2\)
\(\left|1-3x\right|=\dfrac{-7}{30}\)
\(\left|1-3x\right|\in N\) nhưng \(\dfrac{-7}{30}\notin N\)
\(\Rightarrow x=\varnothing\)
b) \(\left(2,8x+50\right):\dfrac{-3}{2}=51\)
\(\left(2,8x+50\right)=51.\dfrac{-3}{2}\)
\(2,8x+50=\dfrac{-153}{2}\)
\(2,8x=\dfrac{-153}{2}-50\)
\(2,8x=\dfrac{-253}{2}\)
\(x=\dfrac{-253}{2}:2,8\)
\(x=\dfrac{-1265}{28}\)
c) \(\dfrac{x-2}{-2}=\dfrac{x+4}{3}\)
\(\Rightarrow\left(x-2\right).3=-2.\left(x+4\right)\)
\(x.3-2.3=\left(-2\right).x+\left(-2\right).4\)
\(3x-6=\left(-2\right)x+\left(-8\right)\)
\(3x-\left(-2\right)x=6+\left(-8\right)\)
\(5x=-2\)
\(x=\left(-2\right):5\)
\(x=\dfrac{-2}{5}\)
d) \(4\left(3-2x\right)-5\left(x-1\right)=12\)
\(4.3-4.2x-5x+5.1=12\)
\(12-8x-5x+5=12\)
\(12+\left(-8\right)x+\left(-5\right)x+5=12\)
\(12+\left(-13\right)x+5=12\)
\(\left(-13\right)x=12-12-5\)
\(\left(-13\right)x=-5\)
\(x=\left(-5\right):\left(-13\right)\)
\(x=\dfrac{5}{13}\)
Q
14 tháng 4 2017
Bài 2: Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\) (đpcm)
23 tháng 8 2022
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
DA
12 tháng 7 2017
a) \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2.3\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{\left(0,2\right)^5.3^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{3^5}{0,2}=\dfrac{243}{0,2}=1215\)
c) \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)
30 tháng 3 2017
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{2}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{-2}{3}\)
\(x=\dfrac{1}{10}.\dfrac{-3}{2}\)
\(x=\dfrac{-3}{20}\)
N
27 tháng 6 2018
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
TH
27 tháng 6 2018
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
2 tháng 5 2017
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
\(\Rightarrow\)\(\left[3x+\dfrac{3}{2}+3-5x\right].\dfrac{1}{2}=-x:\dfrac{31}{5}\)
\(\Rightarrow\)\(\left[-2x+\dfrac{9}{2}\right].\dfrac{1}{2}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(\dfrac{-2x}{2}+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(-x+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5}{31}+x\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5+31x}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x\left(-5+31\right)}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x.26}{31}\)
\(\Rightarrow\dfrac{9}{4}.31=x.26\)\(\Rightarrow\dfrac{279}{4}=x.26\)
\(\Rightarrow x=\dfrac{279}{4}:26=\dfrac{279}{4}.\dfrac{1}{26}\)
\(\Rightarrow x=\dfrac{279}{104}\)
Vậy x=\(\dfrac{279}{104}\)