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

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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

Lời giải:

a.

\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)

\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)

\(=-a^4b^4(3a+4b)^2\)

b.

$x^3-6x^2y+12xy^2-8x^3$

$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$

c.

$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$

$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$

$=(x+\frac{1}{2})^3$

a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

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

b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)

\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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

c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=\left(x+\dfrac{1}{2}\right)^3\)

`a, 1/2 +x=3/4`

`=> x= 3/4 -1/2`

`=> x= 3/4-2/4`

`=>x= 1/4`

`b, 5/2 -x=1/3`

`=> x= 5/2 -1/3`

`=> x= 15/6 - 2/6`

`=>x= 13/6`

`c, 2 . (1/3 +x)=1/5`

`=> 1/3 +x=1/5:2`

`=> 1/3 +x= 1/10`

`=>x= 1/10-1/3`

`=>x= 3/30 - 10/30`

`=>x=-7/30`

`d, 2/3 - (1/2 -x)=1/5`

`=> 1/2-x= 2/3 -1/5`

`=>1/2-x= 10/15 - 3/15`

`=>1/2-x=7/15`

`=>x= 1/2-7/15`

`=>x=1/30`

`1/2 + x = 3/4`

`=>    x  = 3/4 - 1/2`

`=>    x   = 1/4`

`5/2 - x  = 1/3`

`=>    x  =  5/2 - 1/3`

`=>    x  = 13/6`

`2.(1/3 + x) = 1/5`

`=>1/3 + x  = 1/10 `

`=>         x =  1/10 - 1/3`

`=>        x   = -7/30`

`2/3 - (1/2 -x)= 1/5`

`=>     1/2 - x = 7/15`

`=>             x  = 1/2 - 7/15`

`=>             x  = 1/30`

a. \(\dfrac{1}{2}+x=\dfrac{3}{4}\)

⇔ \(x=\dfrac{3}{4}-\dfrac{1}{2}\)

⇔ \(x=\dfrac{1}{4}\)

b. \(\dfrac{5}{2}-x=\dfrac{1}{3}\)

⇔ \(-x=\dfrac{1}{3}-\dfrac{5}{2}\)

⇔ \(-x=-\dfrac{13}{6}\)

⇔ \(x=\dfrac{13}{6}\)

c. \(2\left(\dfrac{1}{3}+x\right)=\dfrac{1}{5}\)

⇔ \(\dfrac{1}{3}+x=\dfrac{1}{5}\div2\)

⇔ \(x=\dfrac{1}{10}-\dfrac{1}{3}\)

⇔ \(-\dfrac{7}{30}\)

d. \(\dfrac{2}{3}-\left(\dfrac{1}{2}-x\right)=\dfrac{1}{5}\)

⇔ \(-\dfrac{1}{2}+x=\dfrac{1}{5}-\dfrac{2}{3}\)

⇔ \(x=-\dfrac{7}{15}+\dfrac{1}{2}\)

⇔ \(x=\dfrac{1}{30}\)

a: \(A=\dfrac{x+2}{x}\cdot\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

b: \(B=\dfrac{2x-6+3x+9-5x+2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5}{x^2-9}\)

a) \(\dfrac{29}{60}\)

b)\(1\dfrac{23}{25}\)

c) \(2\dfrac{26}{35}\)

x = 4 x 3/4

x = 3

x = 4/5 + 2/5

x = 6/5

b) /2x-3/-x=5

+) 2x-3>0⇔x>\(\dfrac{3}{2}\)

    2x-3-x=5

⇔2x-x=5+3

⇔x=8 

+) 2x-3<0⇔x<\(\dfrac{3}{2}\)

   -(2x-3)-x=5

⇔-2x+3-x=5

⇔-2x-x=5-3

⇔-3x=2

⇔x=\(\dfrac{-2}{3}\)

    S={8,\(\dfrac{-2}{3}\)}

c)

ĐKXĐ: \(x\notin\left\{1;-3\right\}\)

Ta có: \(\dfrac{2x-1}{x-1}-\dfrac{2x+5}{x+3}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{2x^2+6x-x-3}{\left(x-1\right)\left(x+3\right)}-\dfrac{2x^2-2x+5x-5}{\left(x+3\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

Suy ra: \(2x^2+5x-3-2x^2-3x+5=4\)

\(\Leftrightarrow2x+2=4\)

\(\Leftrightarrow2x=2\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

a: =>-12<x<2y<-9

=>x=-11; y=-5

b: =>-7<3(x-1)<8

\(\Leftrightarrow3\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)

\(\Leftrightarrow x-1\in\left\{2;1;0;-1;-2\right\}\)

hay \(x\in\left\{3;2;1;0;-1\right\}\)