2​9.19 -29.13-19.29+19.13=(29.19-19.29)+13.(19-29)=0+13.(-10)=-130. Tìm x +2 =1 => x=1-2=-1

a. c1:15x12-3x3x10 = 180-150=30

c2: 15x12-15x10 = 15x(12-10) = 15x2 = 30

b. c1: 45-9x18=45-162=-117

c2: 45-9x13-9x5=45-45-9x13=-117

c. c1: 29x6-19x16=174-304=-130

c2: 29x19-29x13-19x29+19x13=13x(19-29)=13x(-10)=-130

a. 

Cách 1: 

15.12 - 3.5.10

= 180 - 150

= 30

Cách 2:

15.12 - 3.5.10

= 15 .12 - 15.10

= 15. ( 12 -10)

= 15.2

= 30 

Tiếp tương tự. Chúc bạn học tốt!!

a) =180 - 15 x 10

    = 180 - 150 

    = 30

C2 ) 15 x 12 - 15 x 10

= 15 ( 12 - 10 ) 

=15 x 2

=30

b)= 45 - 162

=-117

C2) = 45 - (9 x 13 ) + ( 9 x 5 )

= 45 - 162

= -117

Bài 2 

a: =>x=-23-7=-30

b: =>75:x=15

hay x=5

c: =>2(x+4)=80

=>x+4=40

hay x=36

d: \(\Leftrightarrow2^x\cdot11=88\)

hay x=3

Bài 3: 

Có thể chia được nhiều nhất 12 tổ vì UCLN(180;132)=12

Khi đó, mỗi tổ có 15 nam và 11 nữ

Câu 1:

13*(12-29)-12*(13-29)

=13*12-13*29-12*13+12*29

=-13*29+12*29

=29(-13+12)=29*(-1)=-29

Câu 2:

a) x-2=-1

hay x=1

Vậy: x=1

b) \(-13\cdot\left|x\right|=-26\)

\(\Leftrightarrow\left|x\right|=\frac{-26}{-13}=2\)

\(\Leftrightarrow x\in\left\{2;-2\right\}\)

Vậy: \(x\in\left\{2;-2\right\}\)

c) (x-1)+(x-3)+(x-5)+(x-7)+(x-9)=0

\(\Leftrightarrow x-1+x-3+x-5+x-7+x-9=0\)

\(\Leftrightarrow5x-25=0\)

\(\Leftrightarrow5x=25\)

hay x=5

Vậy: x=5

Câu 1 : Tính : 13 . (12 - 29) - 12 . (13 - 29)

13 . (12 - 29) - 12 . (13 - 29)

= (13 . 12 - 13 . 29) - (12 . 13 - 12 . 29)

= 13 . 12 - 13 . 29 - 12 . 13 + 12 . 29

= (13 . 12 - 12 . 13) + (12 . 29 - 13 . 29)

= (13 . 12 - 12 . 13) + 29 . (12 - 13)

= 0 + 29 . (-1)

= 29 . (-1)

= -29

Câu 2: Tìm số nguyên x biết :

A/ x - 2 = -1

x = (-1) + 2

x = 1

B/ -13 . |x| = -26

|x| = (-26) : (-13)

|x| = 2

x = 2 hay x = -2

C/ ( x - 1 ) + (x - 3 ) + (x - 5 ) + ( x - 7 ) + ( x - 9 ) = 0

x - 1 + x - 3 + x - 5 + x - 7 + x - 9 = 0

(x + x + x + x + x) + [(-1) + (-3) + (-5) + (-7) + (-9)] = 0

5x + (-25) = 0

5x = 0 + 25

5x = 25

x = 25 : 5

x = 5

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

nhiều thế

sfdhg 

\(4^{x-5}=16\)

\(4^{x-5}=4^2\)

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

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

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

a, 45 x 46 + 45 x 45 - 30 - 15= 45x( 46+45) -30 -15= 45 x 91 - 30 - 15= 4095 - 30 - 15= 4050

b, 891 x 29 + 29 x 8+ 29= 29 x ( 891 + 8+ 1)= 29 x 900= 26100

c, 73 x ( 84+17) + 27 x (74+27)= 73 x 101 + 27 x 101= 101 x ( 73+27)= 101 x 100= 10100

d, chịu

d, đề bài= 13 x ( 99 + 2) + 13 x 1001+ 11 x 1300= 13 x 101 + 13 x 1001+ 13 x 1100= 13 x ( 101 + 1001 + 1100) = 13 x 2202= 28626

1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )

= 29 x 6 - 19 x 16

= 174 - 304

=  - 130

2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

⁼ 1 - \(\frac{1}{100}\)

= \(\frac{99}{100}\)