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a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)
=>-10<=x<=-13/7
hay \(x\in\left\{-10;-9;...;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=11/18
hay \(x\in\left\{-1;0\right\}\)
a: \(\Leftrightarrow\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)
=>10<=x<=-13/7
hay \(x\in\varnothing\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=22/36
hay \(x\in\left\{-1;0\right\}\)
\(\Leftrightarrow\dfrac{9}{12}-\dfrac{10}{12}< =\dfrac{x}{12}< =\dfrac{12}{12}-\dfrac{8}{12}+\dfrac{3}{12}\)
=>-1<=x<=7
mà x là số nguyên
nên \(x\in\left\{-1;0;1;2;...;6;7\right\}\)
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{3}{2}\le x\le\dfrac{37}{24}-\dfrac{3-16}{24}=\dfrac{37-3+16}{24}=\dfrac{50}{24}=\dfrac{25}{12}\)
=>3/2<=x<=25/12
mà x là số nguyên
nên x=2
b: \(\Leftrightarrow-\dfrac{1}{23}-\dfrac{3}{23}-\dfrac{7}{23}< x\le\dfrac{1}{23}-\dfrac{8}{23}\)
=>-11<x<=-7
mà x là số nguyên
nên \(x\in\left\{-10;-9;-8;-7\right\}\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
=> 2(2x+1) = 6.7
4x+2=42
4x=40
x=10
Vậy x=10
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)
b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)
d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)
f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)
\(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a-b}\)
=>\(\dfrac{b-a}{ab}=\dfrac{1}{a-b}\)
=>\(-\left(a-b\right)^2=ab\)
Vì a;b>0 nên ab>0
=>\(\left(a-b\right)^2=-ab\)
Mà -ab<0 ;\(\left(a-b\right)^2\)lớn hơn bằng 0 nên
Ko tìm đc gtri nào của a;b thỏa mãn đề bài
a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{-5}{48}\)
\(\dfrac{-1}{12}< x< \dfrac{1}{8}\) hay \(-0,08333...< x< 0,125\)
Vì \(x\in Z\Rightarrow x\in\left\{0\right\}\)
`#3107`
a)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)
Vậy, \(x=-\dfrac{1}{5}\)
b)
\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)
Vậy, \(x=\dfrac{147}{20}\)
c)
\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)
Vậy, \(x=-\dfrac{1}{3}.\)
\(#Emyeu1aithatroi...\)
(2/5 + 3/4 . x)= 11/12 -2/3
(2/5 +3/4 . x)= 1/4
3/4 . x = 1/4 - 2/5
3/4 . x = -3/20
x = -3/20 : 3/4
x = -1/5
Vậy .....
\(-4\dfrac{3}{5}\cdot2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}\cdot1\dfrac{6}{15}\)
\(\Leftrightarrow-10\le x\le-\dfrac{91}{25}\)
\(\Leftrightarrow x\in\left\{-9;-8;-7;-6;-5;-4\right\}\)
\(\Rightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}\le x\le-\dfrac{13}{5}\cdot\dfrac{7}{5}\\ \Rightarrow-10\le x\le-\dfrac{91}{25}\\ \Rightarrow-\dfrac{2500}{25}\le x\le-\dfrac{91}{25}\\ \Rightarrow x\in\left\{-\dfrac{2499}{25};-\dfrac{2498}{25};...;-\dfrac{92}{25};-\dfrac{91}{25}\right\}\)