\(-12\left(x-5\right)+7\left(3-x\right)=15\)

\(-12x+60+21-7x=15\)

\(-12x-7x=15-60-21\)

\(-19x=-66\)

\(x=\frac{66}{19}\)

Vây....

b)

a) -12.(x-5)+7(3-x)=15

<=> - 12x + 60 + 21 - 7x = 15

<=> 12x + 7x = 60 + 21 - 15

<=> 19x = 66

<=> x = \(\frac{66}{19}\)

Vậy x = \(\frac{66}{19}\)

b) 30.(x+2)-6.(x-5)-24.x=100

<=> 30x + 60 - 6x + 30 - 24x = 100

<=> 0x = 10(vô lí)

VẬY x = \(\varnothing\)

-12.(x-5)+7(3-x)\(=\)15

\(\Leftrightarrow\) -12x+60+21-7x\(=\)15

-19x+81=15

-19x=15-81

19x=66

x=66/19

a/ \(-12\left(x-5\right)+7\left(3-x\right)=5\)

\(< =>-12x+60+21-7x=5\)

\(< =>-19x+81=5\)

\(< =>-19x=-76\)

\(< =>x=\frac{76}{19}\)

b/ 30(x+2)-6(x-5)-24x=100

<=>30x + 60 - 6x + 30 - 24x =100

<=> 90=100( vô lý)

c/ \(\left(x-1\right)\left(x^2+1\right)=0\)

\(< =>\hept{\begin{cases}x-1=0\\x^2+1=0\end{cases}}< =>\hept{\begin{cases}x=1\\x^2=-1\left(voly\right)\end{cases}}\)

d/ làm rồi mà

a. \(-12.\left(x-5\right)+7.\left(3-x\right)=5\)

             \(-12x+60+21-7x=5\)

                                    \(-19x+81=5\)

                                                \(-19x=-76\)

                                                         \(x=4\)

b. \(30.\left(x+2\right)-6.\left(x-5\right)-24x=100\)

            \(30x+60-6x+30-24x=100\)

\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)

                                                                 \(90=100\)(vô lí)

                                                              \(\Rightarrow x=\varnothing\)

c. \(\left(x-1\right)\left(x^2+1\right)=0\)

 \(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x^2=-1\left(loại\right)\end{cases}}}\)

 \(\Rightarrow x=1\)

Câu d) chính là câu a) :D

a) - 12 . ( x - 5) + 7 . ( 3 - x ) =5

-12x-(-60)+21-7x=5

-12x+60+21-7x=5

-12x-7x=5-60-21

-19x=-76

19x=76

   x=76:19

   x=4

vậy x=4

b) 30 .( x+ 2 ) - 6 . ( x- 5) -24 .x = 100

30x-60-6x+30-24x=100

(30x-6x-24x)+(60+30)=100

                       90=100 (vô lí)

vậy \(x\in\varnothing\)

c) ( x - 1 ) . ( x ²+ 1) = 0

* x-1=0                  * x²+1=0

      x=0+1                x²=0-1

      x=1                    x²=-1 ( vô lí ,x²  \(\ge0\)⁾

a, => -12x+60+21-7x=5

=> 71-19x=5

=> 19x = 71-5 = 66

=> x = 66 : 19 = 4

b, => 30x+60-6x+30-24=100

=> 24x+76 = 100

=> 24x = 100-76 = 24

=> x = 24 : 24 = 1

Tk mk nha

bài này dễ lắm ....

dù mình ko làm dc

a,        -12x-(-60)+21-7x=5

-19x+81=5

-19x=5-81

-19x=-76

x=-4

b,          30(x+2)-6(x-5)-24=100

30x+60-6x+30-24=100

36x+76=100

36x=34

x=34/36

a. -12.(x-5)+7.(3-x)=5

=> -12x+60+21-7x=5

=> -19x+81=5

=> -19x=5-81

=> -19x=-76

=> x=(-76):(-19)

=> x=4

b. 30.(x+2)-6.(x-5)-24.x=100

=> 30x+60-6x+30-24x=100

=> (30x-6x-24x)+(60+30)=100

=> 0x+90=100

=> 0x=100-90

=> 0x=10 (vô lí)

=> không tồn tại x.

a) \(\left(x-3\right)^2\) = \(\left(-3\right)^2\) + \(\left(-4\right)^2\)

\(\left(x-3\right)^2=9+16\)

\(\left(x-3\right)^2=25\)

\(\left(x-3\right)^2=5^2\)

\(\Rightarrow x-3=5\)

\(\Rightarrow x=5+3=8\)

Vậy x = 8

b) -12 . (x - 5) +7 . (3 - x) = 5

-12x + 60 + 21 - 7x = 5

-12x - 7x = 5 - 60 - 21

-19x = -76

x = -76 : (-19) =4

Vậy x = 4

c) 30 . (x + 2) - 6 . ( x - 5) - 24x = 100

30x + 60 - 6x + 30 - 24x =100

30x - 6x - 24x = 100 - 30 - 60

0x = 10

x=10 : 0

\(\Rightarrow x\in\varnothing\)

a,-12.(x-5)+7.(3-x)=5

=>-12x-(-60)+21+7x=5

=>-12x+60+21+7x=5

=>-12x+7x=5-60-21

=>-5x=-76

=>x=-76:-5

=>x=15,2

b,30.(x+2)-6.(x-5)-24.x=100

=>30.x+60-(6.x-30)-24.x=100

=>30.x+60-6.x+30-24x=100

=>30.x-6.x-24.x=100-60-30

=>0x=10(vô lý)

Vậy không có x thỏa mãn

a, x=15,2

b, x (ko có)

duyệt đi olm