a: =>-12x=60-12=48

=>x=-4

b: =>-5x=24+6-5=25

=>x=-5

Chi tiết nha

Lời giải:
a.

$(-12).x=(-15)(-4)-12$

$(-12)x=60-12=48$

$x=48:(-12)=-4$

b.

$(-5)x+5=(-3)(-8)+6$

$(-5)x+5=24+6=30$

$(-5)x=30-5=25$

$x=25:(-5)=-5$

\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

\(a,\left(8+x\right)\left(6-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\\ b,x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a) (8+x).(6-x)=0
<=> 8+x = 0 hoặc 6-x = 0
=> x = -8 hoặc x = 6

b) c) x^2 - 5x=0
<=> x^2 = 0 hoặc -5x = 0
=> x = 0 hoặc x = 5

a) [8+x]x[6-x]=0

vì tích của chúng phải bằng 0 nên có 2 trường hợp

TH1:(8+(-8))x(6-(xEN))

TH2:(8+(XEN))x(6-6)

x = 8

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) Ta có: \(\dfrac{x}{7}=\dfrac{6}{21}\)

nên \(x=\dfrac{6\cdot7}{21}=\dfrac{42}{21}=2\)

b) Ta có: \(\dfrac{-5}{y}=\dfrac{20}{28}\)

nên \(y=\dfrac{-5\cdot28}{20}=\dfrac{-140}{20}=-7\)

c) Ta có: \(\dfrac{-4}{8}=\dfrac{-7}{y}\)

nên \(y=\dfrac{-7\cdot8}{-4}=\dfrac{-56}{-4}=14\)

❤                                              ✔

a) Tìm số nguyên x, biết:

Với \(x\in Z\), ta có:

9 - x = 8 - (2x + 16)

<=> 9 - x = 8 - 2x - 16 

<=> 9 - x = -2x - 8

<=> x  = -17 (TM)

Vậy x = -17

b) Với \(x\in Z\), ta có:

18 - 2x = 21 - (3x - 5)

<=> 18 - 2x = 21 - 3x + 5

<=> 18 - 2x = 26 - 3x

<=> x = 8 (TM)

Vậy x = 8

a) Ta có: 9-x=8-(2x+16)

\(\Leftrightarrow9-x=8-2x-16\)

\(\Leftrightarrow9-x=-2x-8\)

\(\Leftrightarrow9-x+2x+8=0\)

\(\Leftrightarrow x+17=0\)

hay x=-17

Vậy: x=-17

b) Ta có: 18-2x=21-(3x-5)

\(\Leftrightarrow18-2x=21-3x+5\)

\(\Leftrightarrow18-2x+3x-26=0\)

\(\Leftrightarrow x-8=0\)

hay x=8

Vậy: x=8

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,-4/7=x/21
-12/21 = x/21
x= -12 

b,(x-3)/15=1/-5
x - 3 = -1/5 * 15
x - 3 = -3
x = 0

c,.(3x+8)/-12=-5/30
=> 3x + 8 = 2
=> 3x=-6
=>x=-2

Lời giải:

a. $-5< x< 4$

$\Rightarrow x\in\left\{-4; -3; -2; -1; 0; 1; 2; 3\right\}$

Tổng các số nguyên $x$: $(-4)+(-3)+(-2)+(-1)+0+1+2+3=-4$

b. $-8\leq x\leq 5$

$\Rightarrow x\in\left\{-8; -7; -6; -5; -4; -3; -2; -1; 0; 1; 2;3 ; 4; 5\right\}$

Tổng các giá trị nguyên của $x$:

$(-8)+(-7)+(-6)+...+0+1+2+3+4+5=-21$

c.

$-99\leq x\leq 99$

$\Rightarrow x\in\left\{-99; -98; -97;...; -1; 0; 1; 2; 3;...;99\right\}$

Tổng các số nguyên x: 

$(-99)+(-98)+(-97)+...+(-1)+0+1+2+...+99$
$=[(-99)+99]+[(-98)+98]+....+[(-1)+1=0+0+0+...+0=0$