Tìm x
1/2 + 3/4 < x < 8/10 : 4/14
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a: \(\dfrac{1}{3}+\dfrac{3}{35}< \dfrac{x}{210}< \dfrac{4}{7}+\dfrac{3}{5}+\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{70}{210}+\dfrac{18}{210}< \dfrac{x}{210}< \dfrac{120}{210}+\dfrac{126}{210}+\dfrac{105}{210}\)
\(\Leftrightarrow88< x< 441\)
b: Ta có: \(\dfrac{5}{3}+\dfrac{-14}{3}< x< \dfrac{8}{5}+\dfrac{18}{10}\)
\(\Leftrightarrow\dfrac{5}{3}-\dfrac{14}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}\)
=>-9/3<x<17/5
=>-3<x<3,4
mà x là số nguyên
nên \(x\in\left\{-2;-1;0;1;2;3\right\}\)
Bài 1:
1: =-5/24+16/27+3/4
=-5/24+18/24+16/27
=13/24+16/27
=117/216+128/216=245/216
2: =-1/3+1/3+6/7=6/7
3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)
4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)
\(f\left(x\right)=x^2-2\left(m+5\right)x+m^2+4m-3=0\)
Phương trình cho có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow6m+28>0\Leftrightarrow m>-\frac{14}{3}\left(1\right)\)
ycbt\(\Leftrightarrow\hept{\begin{cases}-2< m+5< 4\\f\left(-2\right)>0\\f\left(4\right)>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-7< m< -1\\m^2+8m+21>0\\m^2-4m-27>0\end{cases}}\Leftrightarrow-7< m< 2-\sqrt{31}\left(2\right)\)
Từ (1),(2) suy ra \(-\frac{14}{3}< m< 2-\sqrt{31}.\)
Câu 1:
\(\frac{1}{3}+\frac{3}{35}<\frac{x}{210}<\frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}<\frac{x}{210}<\frac{158}{105}\)
\(\Rightarrow\frac{88}{210}<\frac{x}{210}<\frac{316}{210}\)
\(\Rightarrow x\in\left\{89;90;91;92;...;310;311;312;313;314;315\right\}\)
Câu 3:
\(\frac{5}{3}\)\(+\frac{-14}{3}\)\(<\)\(x\)\(<\)\(\frac{8}{5}+\frac{18}{10}\)
\(\Rightarrow\)\(-9\)\(<\)\(x\)\(<\)\(3,4\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{-8;-7;-6;-5;...;1;2;3\right\}\)