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a) ta co:
1/18<x/12<y/9<1/4
=>2/36<x.3/36<y.4/36<9/36
=>x.3thuộc{3;6};y.4thuộc{4;8}
=>x thuộc{1;2};y thuộc{1:2}
b) ta co
7/8<x/40<9/10
=>70/80<x.2/40<72/80
=>x.2 =71
=>x=71/2
)
bạn cứ tính 2 vế là xong mà:
a) x\(\in\){1;2;3;4;5;6;7}
b) x=0
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 \(x\in\left\{-8;-7\right\}\)
a: \(\dfrac{7}{11}< x-\dfrac{1}{7}< \dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{7}{11}+\dfrac{1}{7}< x< \dfrac{10}{13}+\dfrac{1}{7}\)
hay 60/77<x<83/91
b: \(\dfrac{7}{9}< \dfrac{13}{11}-x< \dfrac{15}{16}\)
\(\Leftrightarrow\dfrac{-7}{9}>x-\dfrac{13}{11}>-\dfrac{15}{16}\)
\(\Leftrightarrow-\dfrac{7}{9}+\dfrac{13}{11}>x>\dfrac{-15}{16}+\dfrac{13}{11}\)
\(\Leftrightarrow\dfrac{40}{99}>x>\dfrac{43}{176}\)
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
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 ...
\(\dfrac{-7}{12}\) < \(\dfrac{x}{46}\) < \(\dfrac{-8}{-15}\)
\(\dfrac{-7}{12}\) < \(\dfrac{x}{46}\) < \(\dfrac{8}{15}\)
12 = 22.3; 46 = 2.23; 15 = 3.5 => MCNN = 22.3.23.5 = 1380
\(\dfrac{-805}{1380}\) < \(\dfrac{30x}{1380}\) < \(\dfrac{736}{1380}\)
-805 < 30\(x\) < 736
\(-\dfrac{805}{30}\) < \(x\) < \(\dfrac{736}{30}\)
\(\dfrac{-161}{6}\) < \(x\)< \(\dfrac{368}{15}\)
Không có điều kiện gì à bạn