tìm XEZ biết
a) \(\dfrac{3}{8}\) < \(\dfrac{x}{4}\) < \(\dfrac{3}{4}\)
b) \(\dfrac{5}{8}\) < \(\dfrac{x}{6}\) ≤ \(\dfrac{5}{10}\)
c) \(\dfrac{-8}{15}\) < \(\dfrac{x}{40}\) < \(\dfrac{-7}{15}\)
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b ) 3 - ( x - 5 ) = - 7 + ( - 8 )
3 - ( x - 5 ) = - 15
x - 5 = 3 - ( - 15 )
x - 5 = 18
x = 18 + 5
x = 23
Vậy x = 23
a) -15÷x=3
x=-15÷3
x=-5
b) -3x+8=-7
-3x . =-15
x. =-15÷-3
x. = 5
c) ( x-6) (7- x) =0
Suy ra: * x-6 =0=) x=6
*7-x=0=) x=7
Vậy x=6;7
a, -15:x=3
\(\Rightarrow x=-15:3\)
\(\Rightarrow x=-5\)
Vậy x=-5
b,-3x+8=-7
\(\Rightarrow-3x=-7-8\)
\(\Rightarrow-3x=15\)
\(\Rightarrow x=15:\left(-3\right)\)
\(\Rightarrow x=5\)
Vậy x=5
c, (x-6)(7-x)=0
\(\Rightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+6\\x=7-0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)
\(\Rightarrow2x=\dfrac{4}{3}+\dfrac{2}{3}\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)
\(\Rightarrow5:x=\dfrac{5}{8}+\dfrac{3}{8}\)
\(\Rightarrow5:x=1\)
\(\Rightarrow x=5:1=5\)
a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)
\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)
\(\Rightarrow-5:x=\dfrac{5}{8}-\dfrac{-3}{8}\)
\(\Rightarrow-5:x=1\)
\(\Rightarrow x=-5\)
a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)
\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow2x=\dfrac{6}{3}\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=\dfrac{2}{2}\)
\(\Rightarrow x=1\)
b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)
\(\Rightarrow-5:x=\dfrac{-3}{8}-\dfrac{5}{8}\)
\(\Rightarrow-5:x=-\dfrac{8}{8}\)
\(\Rightarrow-5:x=-1\)
\(\Rightarrow x=-5:-1\)
\(\Rightarrow x=5\)
a) 6 ⋮ (x - 1)
=> x - 1 ϵ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
TH1: x - 1 = -6 => x = -5 (Thỏa mãn)
TH2: x - 1 = -3 => x = -2 (Thỏa mãn)
TH3: x - 1 = -2 => x = -1 (Thỏa mãn)
TH4: x - 1 = -1 => x = 0 (Thỏa mãn)
TH5: x - 1 = 1 => x = 2 (Thỏa mãn)
TH6: x - 1 = 2 => x = 3 (Thỏa mãn)
TH7: x - 1 = 3 => x = 4 (Thỏa mãn)
TH8: x - 1 = 6 => x = 7 (Thỏa mãn)
Vậy x ϵ {-5; -2; -1; 0; 2; 3; 4; 7}
b) (x + 2) ⋮ (x - 1)
Ta có: (x + 2) = (x - 1) + 3
Vì (x - 1) ⋮ (x - 1) nên để (x - 1) + 3 ⋮ (x - 1) thì 3 ⋮ (x - 1)
=> x - 1 ϵ Ư(3) = {-3; -1; 1; 3}
TH1: x - 1 = -3 => x = -2 (Thỏa mãn)
TH2: x - 1 = -1 => x = 0 (Thỏa mãn)
TH3: x - 1 = 1 => x = 2 (Thỏa mãn)
TH4: x - 1 = 3 => x = 4 (Thỏa mãn)
Vậy x ϵ {-2; 0; 2; 4}
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