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các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
Bài 2:a)<=>5(x+4) =125-38
<=>5(x+4)=87
<=>x+4=17,4
<=>x=17,4-4=13.4
b)<=> ( 3x – 24) . 73 = 2.74
<=> (3x – 16) = 2.74 : 73
<=> (3x – 16) =2.7
<=>3x – 16 = 14
<=> 3x = 30
<=> x = 10
c)=> xϵ ƯC (70, 84) và x > 8
Ta có: 70 = 2. 5. 7 => ƯCLN( 70,84) = 2. 7 = 14
84 = 22. 3. 7
=> ƯC (70, 84) = Ư(14) = { 1 ;2; 7; 14 }
Vì x > 8 => x = 14
d)=>xϵ BCNN (12, 25, 30) và 0 < x < 500
=>BC(12,25,30)=B(300) = { 0; 300; 600; …}
Vì 0 < x < 500
=> x = 300
giup minh nha cac ban . lam duoc minh se tick cho va ket ban . cam on cac ban
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
a,
=>x€{-19;-18;-17;-16;...;20}
=20
b,=>x€{-18;-17;-16;-15;-14;...;17}
x-18
c,
=>x€{-26;-25;-24;-23;-22;...;26}
=0
d
x€{3;2;1;0;-1;-2;-3;......}
=>|x|={3;1;2}
=>=6
e,x€{4;3;2;1;0;-1;-2;-3;-4;...}
=>|x|€{4;3;2;1;0}
=>=10
1) \(\frac{4}{5}.x=\frac{8}{35}\Rightarrow x=\frac{8}{25}:\frac{4}{5}\Rightarrow x=\frac{2}{5}\)
\(\frac{2}{3}.x+\frac{1}{4}=\frac{7}{12}\Rightarrow\frac{2}{3}.x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}.x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:\frac{2}{3}\)
\(x=\frac{1}{2}\)
\(\frac{x}{10}=\frac{3}{10}\Rightarrow x=\frac{10.3}{10}\Rightarrow x=3\)
a) Ta có: \(\frac{-6}{12}< \frac{x}{12}< \frac{-4}{12}\Rightarrow-6< x< -4\Rightarrow x=-5\)
b) \(\frac{-6}{5}+\frac{13}{15}< \frac{x}{15}< \frac{-2}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{-1}{3}< \frac{x}{15}< \frac{-1}{15}\)
\(\Rightarrow\frac{-5}{15}< \frac{x}{15}< \frac{-1}{15}\)
\(\Rightarrow-5< x< -1\)
\(\Rightarrow x=\left\{-4;-3;-2\right\}\)