a,3/4 . (-5/12)+3/4.(-7/12)

` 3/4 . [ - ( 5/12 + 7/12  ) ] `

`3/4 . (-1) = -3/4 `

`2/3 . x - 0,5 = 3/4 `

`        x - 0,5 = 3/4 - 2/3 `

`        x-0,5 = 1/12 `

`        x =       1/12 + 0,5 `

`        x= 7/12 `

a, 3/4 . (-5/12)+3/4.(-7/12) 

= 3/4 . [(-5/12) + (-7/12)]

= 3/4 . (-1)

= -3/4

----------------------------------------------------------------------------

a, 2/3 .x-0,5=3/4

2/3 . x = 3/2 + 0,5

2/3 . x = 2 

x = 2 : 2/3

x = 3 

vậy x = 3

Bài 2:

a: =>2/3x=3/4+1/2=3/4+2/4=5/4

=>x=5/4:2/3=5/4*3/2=15/8

b:=>-2x+4=3x-12

=>-5x=-16

=>x=16/5

\(a,x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ b,x^2-2x-15=\left(x^2-5x\right)+\left(3x-15\right)=x\left(x-5\right)+3\left(x-5\right)=\left(x+3\right)\left(x-5\right)\\ c,x^2-4+\left(x-2\right)^2=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\)

\(d,x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\)

4:

a: f(x)=0

=>-x-4=0

=>x=-4

b: g(x)=0

=>x^2+x+4=0

Δ=1^2-4*1*4=1-16=-15<0

=>g(x) ko có nghiệm 

c: m(x)=0

=>2x-2=0

=>x=1

d: n(x)=0

=>7x+2=0

=>x=-2/7

Tìm x:

\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)

\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)

Đáp án là C

a)

`x/(-8)=5/4`

`=>x*4=5*(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

b)

`(-13)/x=(-65)/10`

`=>(-13)/x=(-13)/2`

`=>x=2`

c)

`(-6)/5=(-12)/x`

`<=>(-12)/10=(-12)/x`

`<=>x=10`

d)

`(x-2)/2=8/(x-2)`

`=>(x-2)(x-2)=8*2`

`=>(x-2)^2=16`

`=>x-2=4` hoặc`x-2=-4`

`=>x=4+2` hoặc `x=-4+2`

`=>x=6` hoặc `x=-2`

`x/(-8) = 5/4`

`=> 4x=5.(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

__

`-13/x =-65/10`

`=> -65x=-130`

`=>x=-130:(-65)`

`=>x= 2`

__

`-6/5 =-12/x`

`=> -6x=-12.5`

`=> -6x= -60`

`=>x=-60:(-6)`

`=>x=10`

__

`(x-2)/2 = 8/(x-2)`

`=>(x-2)^2=2.8`

`=>(x-2)^2=16`

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

\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

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

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