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c) \(47-\left(x-15\right)=21\)
\(\Leftrightarrow x-15=47-21\)
\(\Leftrightarrow x-15=26\)
\(\Leftrightarrow x=26+15=41\)
d) \(-5-\left(24-x\right)=-11\)
\(\Leftrightarrow24-x=-5+11\)
\(\Leftrightarrow24-x=6\)
\(\Leftrightarrow x=24-6=18\)
a) 16-x=21+8
16-x=29
x=16-29
x=-13
b)x - 32= - 22
x = - 22 +32
x = 10
a) \(\text{16 - x = 21 - (-8).}\)
<=> \(\text{16 - x = 29.}\)
<=> \(x=-13.\)
Vậy \(x=-13.\)
b) \(\text{x - 32 = (-5) - 17}.\)
<=> \(\text{x - 32 = -22}.\)
<=> \(x=10.\)
Vậy \(x=10.\)
a) x + 20 = 15 => x = -5
b)16 + x = -7 => x = -23
c) -8 + x = 13 => x = 21
Câu d hình như sai rồi nha
a) x + 20 = 15
x = 15 - 20
x = -5 (nhận)
Vậy x = -5
b) 16 + x = -7
x = -7 - 16
x = -23 (nhận)
Vậy x = -23
c) -8 + x = 13
x = 13 + 8
x = 21 (nhận)
Vậy x = 21
d) 2 + (-x) = 11
2 - x = 11
x = 2 - 11
x = -9 (nhận)
Vậy x = -9
106 − x = − 58 – 8 106 − x = − 66 − x = − 66 − 106 − x = − 172 x = 172.
Lời giải:
a. $\frac{x}{7}=\frac{6}{21}$
$x=\frac{6}{21}.7$
$x=2$
b.
$\frac{-5}{y}=\frac{20}{28}$
$y=-5:\frac{20}{28}$
$y=-7$
c.
$\frac{-4}{8}=\frac{-7}{y}$
$y=-7:\frac{-4}{8}$
$y=14$
a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)
b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)
c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)
a: =>19/23>19/x>19/29
=>\(x\in\left\{24;25;26;27;28\right\}\)
b: =>88/132<88/x<88/128
=>132>x>128
=>\(x\in\left\{131;130;129\right\}\)
c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)
=>32<x^2<40
=>x=6
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
| x+1 | 1 | -1 | 5 | -5 |
| y-2 | 5 | -5 | 1 | -1 |
| x | 0 | -2 | 4 | -6 |
| y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
| x-5 | 1 | -1 | 7 | -7 |
| y+4 | -7 | 7 | -1 | 1 |
| x | 6 | 4 | 12 | -2 |
| y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
a) x – 32 = − 5 − 17 ⇔ x – 32 = − 32 ⇔ x = − 32 + 32 ⇔ x = 0.
b) 16 – x = 21 – − 8 ⇔ 16 – x = 21 + 8 ⇔ 16 − x = 30 ⇔ − x = 30 − 16 ⇔ − x = 14 ⇔ x = − 14.
c) 106 − x = − 58 – 8 ⇔ 106 − x = − 66 ⇔ − x = − 66 − 106 ⇔ − x = − 172 ⇔ x = 172.