\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)

\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)

\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)

Vậy x = 5/19

\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

Vậy x = 1/2 hoặc x = -6

\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy x = 7 hoặc x = -1

a,

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)

\(\Rightarrow5^{x+3}2=2.5^{11}\)

\(\Rightarrow5^{x+3}=5^{11}\)

\(\Rightarrow x+3=11\)

\(\Rightarrow x=8\)

b, (Check lai xem de sai o dau khong nhe)

\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)

Dat 5x ra ben ngoai

\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)

\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)

\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)

\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)

\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)

\(\Rightarrow5^x\left(5^{-3}\right).9379\)

=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)

Dài thế trời!!!

1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14

2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84

3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9

4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1

5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5

6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3

2/3 x 3/8 + 2/3 x 1/8 + 2/3 x 1/2

= 2/3 x ( 3/8 + 1/8 +1/2 )

= 2/3 x 1 = 2/3

2/3 x 3/8 + 2/3 x 1/8 + 2/3 x 1/2

= 2/3 x ( 3/8 + 1/8 +1/2 )

= 2/3 x 1 = 2/3

a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

\(x-\dfrac{3}{4}=\dfrac{9}{4}\)

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11

\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)

f: =>x*4/7=1/5+2/3=3/15+10/15=13/15

=>x=91/60

g: =>1/3:3x=1/5-2/3=-7/15

=>3x=-5/7

=>x=-5/21

h: =>7/8x=-1/9

=>x=-1/9:7/8=-1/9*8/7=-8/63

j: =>(2,8x-50)=34

=>2,8x=84

=>x=30

1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)

\(=\left(9-x^2\right)\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)

3, \(x^4-5x^2+4\)

Đặt \(x^2=t\left(t\ge0\right)\)ta có : 

\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)

\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

`Answer:`

1. `45+x^3-5x^2-9x`

`=x^3+3x^2-8x^2-24x+15x+45x`

`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`

`=(x+3).(x^2-8x+15)`

`=(x+3).(x^2-5x-3x+15)`

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

2. `x^4-2x^3-2x^2-2x-3`

`=x^4+x^3-3x^3+x^2+x-3x-3`

`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`

`=(x+1).(x^3-3x^2+x-3)`

`=(x+1).[x^3 .(x-3).(x-3)]`

`=(x+1).(x-3).(x^2+1)`

3. `x^4-5x^2+4`

`=x^4-x^2-4x^2+4`

`=x^2 .(x^2-1)-4.(x^2-1)`

`=(x^2-1).(x^2-4)`

`=(x-1).(x+1).(x-2).(x+2)`

4. `x^4+64`

`=x^4+16x^2+64-16x^2`

`=(x^2+8)^2-16x^2`

`=(x^2+8-4x).(x^2+8+4x)`

5. `x^5+x^4+1`

`=x^5+x^4+x^3-x^3+1`

`=x^3 .(x^2+x+1)-(x^3-1)`

`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`

`=(x^2+x+1).(x^3-x+1)`

6. `(x^2+2x).(x^2+2x+4)+3`

`=(x^2+2x)^2+4.(x^2+2x)+3`

`=(x^2+2x)^2+x^2+2x+3.(x^2+2x)+3`

`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`

`=(x^2+2x+1).(x^2+2x+3)`

`=(x+1)^2 .(x^2+2x+3)`

7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`

`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`

`=x^6+8x^4+19x^3+30x^2+88x+64`

8. `x^3 .(x^2-7)^2-36x`

`=x[x^2.(x^2-7)^2-36]`

`=x[(x^3-7x)^2-6^2]`

`=x.(x^3-7x-6).(x^3-7x+6)`

`=x.(x^3-6x-x-6).(x^3-x-6x+6)`

`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`

`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`

`=x.(x+1).(x-1).(x^2-3x+2x-6).(x^2+3x-2x-6)`

`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`

`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`

9. `x^5+x+1`

`=x^5-x^2+x^2+x+1`

`=x^2 .(x^3-1)+(x^2+x+1)`

`=x^2 .(x-1).(x^2+x+1)+(x^2+x+1)`

`=(x^2+x+1).(x^3-x^2+1)`

10. `x^8+x^4+1`

`=[(x^4)^2+2x^4+1]-x^4`

`=(x^4+1)^2-(x^2)^2`

`=(x^4-x^2+1).(x^4+x^2+1)`

`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`

`=[(x^2+1)^2-x^2].(x^4-x^2+1)`

`=(x^2-x+1).(x^2+x+1).(x^4-x^2+1)

11. ` x^5-x^4-x^3-x^2-x-2`

`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`

`=x^4 .(x-2)+x^3 ,(x-2)+x^2 .(x-2)+x.(x-2)+(x-2)`

`=(x-2).(x^4+x^3+x^2+x+1)`

12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`

`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`

`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`

`=(x^2-1).(x^7-x^4-x^3+1)`

`=(x-1)(x+1)(x^3-1)(x^4-1)`

`=(x-1)(x+1)(x^2+x+1)(x-1)(x^2-1)(x^2+1)`

`=(x-1)^2 .(x+1)(x^2+x+1)(x-1)(x+1)(x^2+1)`

`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`

13. `(x^2-x)^2-12(x^2-x)+24`

`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`

`=(x^2-x+6)^2-12`

`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`

b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)

\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)

\(\Leftrightarrow x+3=11\)

hay x=8

c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)

\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)

\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)

Suy ra: x=5

d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)

\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)

\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)

\(\Leftrightarrow8^x=8^{20}\)

Suy ra: x=20

mình đang cần gấp câu trả lời ,bạn nào giải được nhanh k luôn ,hứa đấy

bạn đi học thêm à ?

1) -5+(-3+12-x)=7-(2-8)

=>-5+(9-x)=7-(-6)

=>-5+9-x=13

=>x=9-5-13=-9

2) 8-(7-x+3)=-|-7+2|-(-9)

=>8-(10-x)=-|-5|+9

=>8-10+x=-5+9

=>x=-5+9-8+10=6

3)-(12-5+9-x)=3-|5-8+2|

=>-(16-x)=3-|-1|

=>-16+x=3-1

=>x=3-1+16=18

4)-(-3-x+8)-(-3+5)=|5-8|

=>-(5-x)-2=|-3|

=>-5+x-2=3

=>x=3+5+2=10

5)-|5-8|-(4-x+5)=4+|3-9|

=>-|-3|-(9-x)=4+|-6|

=>-3-9+x=4+6

=>x=4+6+3+9=22