tìm số nguyên x :
\(\dfrac{-8}{15}\)<\(\dfrac{x}{40}\)≤\(\dfrac{-7}{15}\)
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a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
Tìm số nguyên x, biết:
a) \(\dfrac{-28}{35}=\dfrac{16}{x};\) b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}.\)
\(a.\)
\(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)
\(b.\)
\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)
\(\Leftrightarrow x=-17\)
a: \(\dfrac{x+2}{27}=\dfrac{x}{-9}\)
=>x+2=-3x
=>4x=-2
hay x=-1/2
b: \(\dfrac{-7}{x}=\dfrac{21}{34-x}\)
=>-7(34-x)=21x
=>34-x=-3x
=>2x=-34
hay x=-17
c: \(\dfrac{-8}{15}< \dfrac{x}{40}< \dfrac{-7}{15}\)
\(\Leftrightarrow-64< 3x< -56\)
hay \(x\in\left\{-21;-20;-19\right\}\)
d: \(\dfrac{-1}{2}< \dfrac{x}{18}< \dfrac{-1}{3}\)
=>-9<x<-6
hay \(x\in\left\{-8;-7\right\}\)
Ta có :
MSC của 15 và 40 là 120
Suy ra -8/15 = -64/120
-7/15 = -56/120
Gọi số cần tìm là x ta có
-64/120 < 3x/120 < -56/120
Vậy 3x có thể là -62 ; -60 ; -58
Ta có \(\dfrac{-8}{15}< \dfrac{...}{40}< \dfrac{-7}{15}\)
=> \(\dfrac{-64}{120}< \dfrac{3...}{120}< \dfrac{-56}{120}\)
=> Vậy số nguyên cần điền vào ô trống là -20; -21; -19
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
Bài 4:
a) \(\dfrac{2.7.13}{26.35}=\dfrac{2.7.13}{13.2.7.5}=\dfrac{1}{5}\)
b) \(\dfrac{23.5-23}{4-27}=\dfrac{23.\left(5-1\right)}{-23}=\dfrac{23.4}{-23}=-4\)
c) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)
\(-\dfrac{7}{x}+\dfrac{8}{15}\) = \(\dfrac{-1}{20}\)
\(\dfrac{7}{x}\) = \(\dfrac{8}{15}\) + \(\dfrac{1}{20}\)
\(\dfrac{7}{x}\) = \(\dfrac{7}{12}\)
\(x\) = 7 : \(\dfrac{7}{12}\)
\(x\) = 12
a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)
Giải:
\(\dfrac{-8}{15}< \dfrac{x}{40}\le\dfrac{-7}{15}\)
\(\Leftrightarrow\dfrac{-8.8}{15.8}< \dfrac{x.3}{40.3}\le\dfrac{-7.8}{15.8}\)
\(\Leftrightarrow\dfrac{-64}{120}< \dfrac{3x}{120}\le\dfrac{-56}{120}\)
\(\Leftrightarrow-64< 3x\le-56\)
\(\Leftrightarrow3x\in\left\{-63;-62;-61;-60;-59;-58;-57;-56\right\}\)
\(\Leftrightarrow x\in\left\{-21;-\dfrac{62}{3};-\dfrac{61}{3};-20;-\dfrac{59}{3};-\dfrac{58}{3};-19;-\dfrac{56}{3}\right\}\)
Mà x ∈ Z
\(\Leftrightarrow x\in\left\{-21;-20;-19\right\}\)
Vậy ...