tìm 2 số nguyên a và b biết :a+b=-1 và a.b=-12.Giup mình nha

1, x^2 . x^3 = 3^7 : 3^2

       x^5     = 3^5

=> x = 3

2, x - 18 : 3 = 12

    x - 6       = 12

    x            = 12 + 6

    x            = 18

3, (x - 18) : 2 = 12

     x - 18       = 12 . 2

     x - 18       = 24

     x             = 24 + 18

     x             = 42

4, x^2 = 16

    x^2 = 4^2

=> x = 4

x=4 nha bn

k k nha

mk k lại cho!

a) \(2\left(x-5\right)-3\left(x+7\right)=14\)

\(\Leftrightarrow2x-10-3x-21=14\)

\(\Leftrightarrow-x-31=14\)

\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)

b) \(5\left(x-6\right)-2\left(x+3\right)=12\)

\(\Leftrightarrow5x-30-2x-6=12\)

\(\Leftrightarrow3x-36=12\)

\(\Leftrightarrow3x=48\Leftrightarrow x=16\)

c) \(3\left(x-4\right)-\left(8-x\right)=12\)

\(\Leftrightarrow3x-12-8+x=12\)

\(\Leftrightarrow4x-20=12\)

\(\Leftrightarrow4x=32\Leftrightarrow x=8\)

d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)

\(\Leftrightarrow-21x+35+14x-28=28\)

\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)

a,|x|+3=5

\(\Leftrightarrow\left|x\right|=5-3=2\)

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

b,|x+3|=5

\(\Rightarrow\left\{{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\\-8\end{matrix}\right.\)

c,|x-7|+13=25

<=>|x-7|=25-13=12

\(\Rightarrow\left\{{}\begin{matrix}x-7=12\\x-7=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=19\\x=-5\end{matrix}\right.\)

d,\(26-\left|x+9\right|=-13\)

\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)=39\)

\(\Rightarrow\left\{{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)

e,8-|x|=15

<=>|x|=8-15=-7

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

f,6-|-3+x|=-15

\(\Leftrightarrow\left|-3+x\right|=6-\left(-15\right)=21\)

\(\Rightarrow\left\{{}\begin{matrix}-3+x=21\\-3+x=-21\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)

a, |x| + 3 = 5

|x| = 5 - 3

|x| = 2

|x| = 2 hoặc |x| = -2

Vậy |x| thuộc {2; -2}

b,|x + 3| = 5

|x + 3| = 5 hoặc |x + 3| = -5

x = 5 - 3 x = (-5) - 3

x = 2 x = -8

Vậy x thuộc {2; -8}

c,|x - 7| + 13 = 25

|x - 7| = 25 - 13

|x - 7| = 12

|x - 7| = 12 hoặc |x - 7| = -12

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

x = 19 x = -5

Vậy x thuộc {19 ; -5}

1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna

Đăng ít thôi

a ) Ta có : 4(x - 5) - 3(x + 7) = -19

<=> 4x - 20 - 3x - 21 = -19

=> x - 41 = -19

=> x = -19 + 41

=> x = 22

b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5

<=> 7x - 21 - 15 + 5x = 11x - 5

<=> 12x - 36 = 11x - 5

=> 12x - 11x = -5 + 36

=> x = 31 

típ đi bạn

Câu a:

1 + {-2 - [-3 + (-4 + |x|)]} = 1 - 2 + [(-3 - 4)]

1 + {-2 -[-3 - 4 + |x|]} = 1 - 2 + [-7]

1 + {-2 - [- 7 + |x|]} = 1 - 2 - 7

1 + {- 2 + 7 - |x|} = - 1 - 7

1 + {5 - |x|} = - 8

5 - |x| = - 8 - 1

5 - |x| = - 9

|x| = 5 + 9

|x| = 14

x = - 14 hoặc x = 14

Vậy x ∈ {-14; 14}

Câu b:

34 + (9 - 21) = 3417 - (x + 3417)

34 + (-12) = 3417 - x - 3417

34 - 12 = 3417 - 3417 - x

22 = - x

x = 22 : (-1)

x = - 22

Vậy x = - 22

Lời giải:

a)

\(-3\frac{5}{8}+\left(-\frac{3}{8}+\frac{9}{4}\right)\)

\(=-\frac{29}{8}+\left(-\frac{3}{8}+\frac{18}{8}\right)\)

\(=-\frac{29}{8}+\frac{15}{8}=-\frac{14}{8}=-\frac{7}{4}\)

b) \(\frac{\left(-9\right)\cdot11+32\cdot\left(-9\right)}{\left(-43\right)\cdot15+12\cdot\left(-43\right)}=\frac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}=\frac{\left(-9\right)\cdot43}{\left(-43\right)\cdot27}=\frac{\left(-1\right)\cdot1}{\left(-1\right)\cdot3}=\frac{1}{3}\)

c) Thay \(x=\frac{2011}{2012}\)vào biểu thức \(x\cdot\frac{1}{3}+2x\cdot\frac{3}{6}-3x\cdot\frac{4}{9}\)ta có :

\(\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{3}{6}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)

\(=\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{1}{2}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)

\(=\frac{2011}{6036}+\frac{2011}{2012}-\frac{2011}{1509}\)

\(=\frac{2011}{6036}+\frac{6033}{6036}-\frac{8044}{6036}=\frac{2011+6033-8044}{6036}=0\)

a: =>15-(x-2)=-13-27=-40

=>x-2=15+40=55

hay x=57

b: =>5-x=-114+12=-102

=>x=107

c: \(\Leftrightarrow\left|x\right|=-1-5=-6\)(vô lý)

d: \(\Leftrightarrow\left|x-3\right|=3\)

=>x-3=3 hoặc x-3=-3

=>x=6 hoặc x=0