\(a,3-\left(17-x\right)=-12\\ \Rightarrow17-x=15\\ \Rightarrow x=2\\ b,-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\\ c,25+\left(-2+x\right)=5\\ \Rightarrow-2+x=20\\ \Rightarrow x=18\\ d,30+\left(32-x\right)=10\\ \Rightarrow32-x=-20\\ \Rightarrow x=52\)

a) `3-(17-x)=-12`

`3-17+x=-12`

`x=-12-3+17`

`x=2`

b) `-26-(x-7)=0`

`-26-x+7=0`

`-19-x=0`

`x=-19`

c) `25+(-2+x)=5`

`25-2+x=5`

`x=5-25+2`

`x=-18`

d) `30+(32-x)=10`

`30+32-x=10`

`62-x=10`

`x=52`

ban chia ra tung bai di dai lam

bai nao lam dc thi giam di nhe

Bài 4:

a) \(\left(x-2\right)\left(y+1\right)=5\)

\(\Leftrightarrow x-2;y+1\inƯ\left(5\right)\)

\(\Leftrightarrow x-2;y+1\in\left\{1;-1;5;-5\right\}\)

*Trường hợp 1:

\(\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)(thỏa mãn)

*Trường hợp 2:

\(\left\{{}\begin{matrix}x-2=5\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=0\end{matrix}\right.\)(thỏa mãn)

*Trường hợp 3:

\(\left\{{}\begin{matrix}x-2=-1\\y+1=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-6\end{matrix}\right.\)(thỏa mãn)

*Trường hợp 4:

\(\left\{{}\begin{matrix}x-2=-5\\y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)(thỏa mãn)

Vậy: x∈{3;7;1;-3} và y∈{4;0;-6;-2}

b) (3-x)*(2y+5)=4

\(\Leftrightarrow3-x;2y+5\inƯ\left(4\right)\)

\(\Leftrightarrow3-x;2y+5\in\left\{1;-1;2;-2;4;-4\right\}\)

*Trường hợp 1:

\(\left\{{}\begin{matrix}3-x=1\\2y+5=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\frac{1}{2}\end{matrix}\right.\)(loại)

*Trường hợp 2:

\(\left\{{}\begin{matrix}3-x=4\\2y+5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

*Trường hợp 3:

\(\left\{{}\begin{matrix}3-x=-1\\2y+5=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\2y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\frac{-9}{2}\end{matrix}\right.\)(loại)

*Trường hợp 4:

\(\left\{{}\begin{matrix}3-x=-4\\2y+5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-3\end{matrix}\right.\)

*Trường hợp 5:

\(\left\{{}\begin{matrix}3-x=2\\2y+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{-3}{2}\end{matrix}\right.\)(loại)

*Trường hợp 6:

\(\left\{{}\begin{matrix}3-x=-2\\2y+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\2y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{-7}{2}\end{matrix}\right.\)(loại)

Vậy: x∈{-1;7} và y∈{-2;-3}

Bài 5:

a) Ta có: \(n+5⋮n-1\)

\(\Leftrightarrow5⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(5\right)\)

\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)

\(\Leftrightarrow n\in\left\{2;0;6;-4\right\}\)(thỏa mãn)

Vậy: \(n\in\left\{2;0;6;-4\right\}\)

b) Ta có: \(2n-3⋮n+4\)

\(\Leftrightarrow-3⋮n+4\)

\(\Leftrightarrow n+4\inƯ\left(-3\right)\)

\(\Leftrightarrow n+4\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow n\in\left\{-3;-5;-1;-7\right\}\)(thỏa mãn)

Vậy: \(n\in\left\{-3;-5;-1;-7\right\}\)

c) Ta có: 3n+4⋮2n-1

\(\Leftrightarrow4⋮2n-1\)

\(\Leftrightarrow2n-1\inƯ\left(4\right)\)

\(\Leftrightarrow2n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow2n\in\left\{2;0;3;-1;5;-3\right\}\)

\(\Leftrightarrow n\in\left\{1;0;\frac{3}{2};\frac{-1}{2};\frac{5}{2};\frac{-3}{2}\right\}\)

Vì n∈Z

nên n∈{1;0}

Vậy: n∈{1;0}

a, 3 - (17 -x ) = -12

       3- 17+ x = 12

        -14       = 12 -  x

      12 - (-14) = x

           2       = x

           x      = 2

a;x=2                                                                                                                                                                             b;x=35                                                                                                                                                                           c;x=-18                                                                                                                                                                         d;x=52                        

giải ra nhé các bn  

a) \(5\left(x-7\right)=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=7\)

b) \(25\left(x-4\right)=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

c) \(\left(34-2x\right)\left(2x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

d) \(\left(2019-x\right)\left(3x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)

e) \(57\left(9x-27\right)=0\)

\(\Rightarrow9x-27=0\)

\(\Rightarrow9\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

a) 5.(x-7)=0⇔x-7=0⇔x=7

b) 25(x-4)=0⇔x-4=0⇔x=4

c) (34-2x).(2x-6)=0

⇔ 34-2x=0 hoặc 2x-6=0

⇔2x=34 hoặc 2x=6

⇔ x=17 hoặc x=3

d) (2019-x).(3x-12)=0

⇔ 2019-x=0 hoặc 3x-12=0

⇔ x=2019 hoặc x=4

e) 57.(9x-27)=0

⇔ 9x-27=0

⇔ x=3

f) 25+(15-x)=30

⇔ 15-x=5

⇔ x=10

g) 43-(24-x)=20

⇔ 24-x=23

⇔ x=1

h) 2.(x-5)-17=25

⇔ 2(x-5)=42

⇔x-5=21

⇔ x=26

i) 3(x+7)-15=27

⇔ 3(x+7)=42

⇔ x+7=14

⇔ x=7

j) 15+4(x-2)=95

⇔ 4(x-2)=80

⇔ x-2=20

⇔ x=22

k) 20-(x+14)=5

⇔ x+14=15

⇔ x=1

l) 14+3(5-x)=27

⇔ 3(5-x)=13

⇔ 5-x=13/3

⇔ x=5-13/3

⇔ x=2/3

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

Bài 1 Tìm x biết:

a)65-(29-x)=32

65 -29+x=31

x=31-65+29

x=-5

b)(x+5)-(x+23)=x-34

x+5 -x +23 = x-34

(x-x)+ (23+5)=x-34

0+28=x-34

28=x-34

28+34=x

62=x

=>x=62

c)(16-x)+(x-38)=x+44

16-x+x-38=x+44

-x+x-x=44-16+38

-x=36

=>x=-36

d)-12+3(-x+7)=-18

3(-x+7)=-18+12

3(-x+7)=-6

-x+7=-6:3

-x+7=-2

-x=-2-7

-x=-9

=>x=9

Baif 2

d)|7-x|=10

=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)

e)(x-6).(7-2x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)

f)(9-x).(2x+8)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

g)x(-x+8).(-3x-18)=0

\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)

h)(-x+8).(x-54).(-24-x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)

1. Thực hiện phép tính bằng cách hợp lí :

a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]

= 0+(-100) = -100

b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75

= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000

c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]

= (-151) + [(-100) + 100] = -151

d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]

= 32+[(-200)+(-400)] = 32+(-600) = -568

e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65

= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)

= 65.(-70) = -4550

g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43

= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)

= 16.(-10) = -160

a)2

b)26

c) -243

d)52

a, => 3-17+x=-12

=> x-14=-12

=> x=-12+14 = 2

b, => 26-x+7=0

=> 33-x=0

=> x=33-0=33

c, => 250-2+x=5

=> x+248=5

=> x=5-248 = -243

d, => 30+32-x=10

=> 62-x=10

=> x=62-10=52

Tk mk nha