Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{7}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}< \frac{x}{210}< \frac{29}{15}\)
\(\Rightarrow\frac{88}{210}< \frac{x}{210}< \frac{406}{210}\)
\(\Rightarrow x\in\left\{89;90;91;...;405\right\}\)
b) \(\frac{5}{3}+-\frac{14}{3}< x< \frac{8}{5}+\frac{8}{10}\)
\(\Rightarrow-3< x< 2\frac{2}{5}\)
=> x thuộc {-2;-1;0;1;2} ( nếu x là số nguyên)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right):\frac{-5}{6}< x< \frac{4}{21}.\frac{4}{7}\)
\(\Rightarrow\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right):\frac{-10}{12}< x< \frac{16}{147}\)
\(\Rightarrow\frac{11}{12}.\frac{-12}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-11}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-1617}{1470}< x< \frac{16}{1470}\)
\(x=\left\{-1;0\right\}\)
Có 1/3+3/35=44/105
Có 4/7+3/5+1/3=158/105
=> 44/105< x/210<158/105 MC: 210
=> 88/210< x/210< 316/210
Vậy x thuộc {89;90;91;92;...;315}
Ta có :
\(\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Leftrightarrow\)\(\frac{70}{210}+\frac{18}{210}< \frac{x}{210}< \frac{120}{210}+\frac{126}{210}+\frac{70}{210}\)
\(\Leftrightarrow\)\(\frac{70+18}{210}< \frac{x}{210}< \frac{120+126+70}{210}\)
\(\Leftrightarrow\)\(\frac{88}{210}< \frac{x}{210}< \frac{316}{210}\)
\(\Leftrightarrow\)\(88< x< 316\)
\(\Rightarrow\)\(x\in\left\{89;90;91;...;314;315\right\}\)
Vậy \(x\in\left\{89;90;91;...;314;315\right\}\)
Chúc bạn học tốt ~
a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...
a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)
\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)
\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)
vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)
mấy câu kia cs tương tự ạ