ta có:

x+2x+3x+...+18x=1368

(1+2+3+...+18)x=1368

x=1368:(1+2+3...+18)

x=1368:171

x=8

=> x(1+2+3+...+18)=1368

=> 171x = 1368

=> x = 8

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a 36x 6+ 36x4

=36x[6+4]

=36x10

=360

b 1368 x99+1368

=1368 x[99+1]

=1368x100

=136800

a) 36x6+36x4=36x(6+4)=36x10=360

b) 1368x99+1368=1368x(99+1)=1368x100=136800

a) 36 x 6 + 36 x 4

= 36 x ( 6 + 4 )

= 36 x 10

= 360

b) 1368 x 99 + 1368

= 1368 x 99 + 1368 x 1

= 1368 x ( 99 + 1 )

= 1368 x 100

= 136 800

a,  70 - 5 ( x - 3 ) = 45

⇔ 5 . x - 3 = 70 - 45

⇔ 5 x - 3 = 35

⇔ x - 3 = 35 : 5 ⇔ x - 3 = 7

⇔ x = 10

b,  10 + 2 x = 4 5 : 4 3

⇔ 10 + 2 x = 4 2 ⇔ 10 + 2 x = 16

⇔ 2 x = 4 ⇔ x = 2

c,  60 - 3 x - 2 = 51

⇔ 3 x - 2 = 60 - 51

⇔ 3 x - 2 = 9

⇔ x - 2 = 3 ⇔ x = 5

d,  4 x - 20 = 2 5 : 2 3

⇔ 4 x - 20 = 2 2 ⇔ 4 x - 20 = 4

⇔ 4 x = 24 ⇔ x = 6

Tách ?

`a, 70 -5.(x-3) =45`

`=> 5.(x-3)= 70-45`

`=> 5.(x-3)=25`

`=>x-3=25:5`

`=>x-3=5`

`=>x= 5+3`

`=>x=8`

______

`b,10 + 2.x = 4^5:4^3`

`=> 10 + 2.x = 4^(5-3)`

`=> 10 + 2.x =4^2=16`

`=> 2.x=16-10`

`=>2.x=6`

`=>x=6:2`

`=>x=3`

_____

`c,60-3.x-2=51`

`=>  60-3.x= 51+2`

`=> 60-3.x=53`

`=>3.x=60-53`

`=> 3.x= 7`

`=>x= 7/3`

____

`d, 4.x-20=2^5:2^3`

`=> 4.x-20=2^(5-3)`

`=> 4.x-20=2^2`

`=> 4.x= 4+20`

`=>4.x=24`

`=>x=24:4`

`=>x=6`

____

`2^x . 4=16`

`=> 2^x=16:4`

`=>2^x= 4`

`=>2^x=2^2`

`=>x=2`

____

`f, 3^x . 3=243`

`=>3^x=243:3`

`=> 3^x=81`

`=> 3^x=3^3`

`=>x=3`

_____

`g, 64. 4^x =16^8`

`=> 4^3 . 4^x=(4^2)^8`

`=> 4^3 . 4^x = 4^(16)`

`=> 4^x= 4^(16-3)`

`=>4^x=4^(13)`

`=>x=13`

_____

`2^x . 16^2 =1024`

`=> 2^x= 1024 : 16^2`

`=>2^x=4`

`=>2^x=2^2`

`=>x=2`

a: =>5(x-3)=25

=>x-3=5

=>x=8

b: =>2x=16-10=6

=>x=3

c: =>58-3x=51

=>3x=7

=>x=7/3

d: =>4x-20=4

=>4x=24

=>x=6

e: =>2^x=4

=>2^x=2^2

=>x=2

f: =>3^x=81

=>3^x=3^4

=>x=4

g: =>4^x*4^3=4^16

=>x+3=16

=>x=13

h: =>2^x=1024/256=4=2^2

=>x=2

a) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow x^3-27+4x-x^3=1\)

\(\Leftrightarrow4x-27=1\Leftrightarrow4x=28\Leftrightarrow x=7\)

b) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

\(\Leftrightarrow x^3-6x^2+12x+4-x^3-8+6\left(x^2-4\right)=60\)

\(\Leftrightarrow-6x^2+12x-4+6x^2-24=60\)

\(\Leftrightarrow12x-28=60\Leftrightarrow x=\frac{22}{3}\)

a) (x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1

x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 + 2x^2 - x^3 + 4x - 2x^2 = 1

4x - 27 = 1

4x = 28

x = 7

b) (x - 2)^3 - (x - 2)(x^2 + 2x + 4) + 6(x - 2)(x + 2) = 60

x^3 - 4x^2 + 4x - 2x^2 + 8x - 8 - x(x^2 + 2x + 4) + 2(x^2 + 2x + 4) + 6x - 24 = 60

x^3 + 12x - 32 - x^2 - 2x^2 - 4x + 2x^2 + 4x + 8 = 60

12x - 24 = 60

12x = 60 + 24

12x = 84

x = 7

Ta có:\(3x=2z-x\Rightarrow4x=2z\Rightarrow2x=z\)

\(x+y+z=60\Rightarrow z=60-x-y\Rightarrow2x=60-x-y\Rightarrow3x=60-y\)

                                                                                                                            \(\Rightarrow4y=60-y\Rightarrow5y=60\Rightarrow y=12\)

\(\Rightarrow4y=3x=12.4=48\Rightarrow x=\frac{48}{3}=16\)

Mà \(2x=z\Rightarrow z=16.2=32\)

Vậy\(x=16;y=12;x=32\)

Ta có :

3x = 4y = 2z => \(\frac{x}{4}=\frac{y}{3}\)và \(\frac{y}{2}=\frac{z}{4}\)=> \(\frac{x}{8}=\frac{y}{6}\)và \(\frac{y}{6}=\frac{z}{12}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có :

\(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}=\frac{x+y+z}{8+6+12}=\frac{60}{24}\)

Suy ra : \(\frac{x}{8}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.8=20\)

               \(\frac{y}{6}=\frac{60}{24}\Rightarrow y=\frac{60}{24}.6=15\)

               \(\frac{z}{12}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.12=30\)

Vậy : x = 20 ; y = 15 ; z = 20 

Bạn ice ❅❅❅❅❅❅ dark sai rồi kết quả của bạn là x=20;y=15;x=30 không có tổng bằng 60 

(2x + 1)(y - 3) = 10

TH1 : 2x + 1 = 1 => 2x = 0 => x = 0

          y - 3 = 10 => y = 13

TH2 : 2x + 1 = 10 => 2x = 9 => x = 4,5

          y - 3 = 1 => y = 4

TH3 : 2x + 1 = 2 => 2x = 1 => x = 0,5

          y - 3 = 5 => y = 8

TH4 : 2x + 1 = 5 => 2x = 4 => x = 2

          y - 3 = 2 => y = 5

TH5 : 2x + 1 = -1 => 2x = -2 => x = -1

          y - 3 = -10 => y = -7

TH6 : 2x + 1 = -10 => 2x = -11 => x = -11/2

          y - 3 = -1 => y = 2 

TH7 : 2x + 1 = -2 => 2x = -3 => x = -3/2

          y - 3 = -5 => y = -2

TH8 : 2x + 1 = -5 => 2x = -6 => x = -3

          y - 3 = -2 => y = 1

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cau nay o biet lam

     \(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)

\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)

\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)

\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)

Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)

Thay vào (1), ta có:

     \(a\left(a+4\right)-60=0\)

\(\Rightarrow a^2+4a-60=0\)

\(\Rightarrow a^2+10a-6a-60=0\)

\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)

\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)

- Nếu \(x^2+7x+6=6\)

\(\Rightarrow x^2+7x=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)

- Nếu \(x^2+7x+6=-10\)

\(\Rightarrow x^2+7x+16=0\)

Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)

Vậy \(x=0,x=-7\)

Học tốt.