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15 tháng 4 2015

    (9/4):(1/3x-1/2)=9/2

=>       (1/3x-1/2)=9/4:9/2=9/4.2/9

=>       (1/3x-1/2)=1/2

=>        1/3x       =1/2+1/2

=>        1/3x       =1

=>            x        =1:1/3=1.3

=>            x        =3

17 tháng 7 2021

(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2

⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)

⇒1-3x2-(9x2-17x-2)= -56x-97

⇒1-3x2-9x2+17x+2=-56x-97

⇒3-12x2+17x=-56x-97

⇒3-12x2+17x+56x+97=0

⇒-12x2+73x+100=0

⇒-(12x2-73x-100)=0

 

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

30 tháng 6 2021

Ta có:(x-3)(x2+3x+9)-x(x2-4)=1

     => x3-27-x3+4x=1

     =>4x=28=>x=7

6 tháng 8 2020

(3x - 4)(x - 2) = 3x(x - 9) - 3

=> 3x2 - 10x + 8 = 3x2 - 27x - 3

=> 27x - 10x = -3 - 8

=> 17x = -11

=> x = -11/17 

\(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Leftrightarrow17x+11=0\)

\(\Leftrightarrow17x=-11\)

\(\Leftrightarrow x=\frac{-11}{17}\)

16 tháng 11 2023

a) 1/4(x-3)+2=1/5

1/4.(x-3) = 1/5-2

1/4.(x-3) = -9/5

x-3 = (-9/5):1/4

x-3 = -36/5

x = -36/5+3

x= -21/5

30 tháng 12 2021

a  x = \(\dfrac{-1}{12}\)

b  x = \(\dfrac{-4}{3}\)

c  x = \(\dfrac{-1}{6}\)

d  x = \(\dfrac{-1}{4}\)

30 tháng 12 2021

\(\left(4x+1\right)^2=\dfrac{4}{9}\)

\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)

\(\text{Vậy }4x+1=\dfrac{2}{3}\)

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

        \(x\)         \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)

\(\text{hoặc }4x+1=\dfrac{-2}{3}\)

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

         \(x\)         \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)

\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)

\(\left(3x-1\right)^2=25\)

\(\left(3x-1\right)^2=\perp\left(5\right)^2\)

\(\text{Vậy }3x-1=5\)

       \(3x\)        \(=5+1=6\)

        \(x\)         \(=6:3=2\)

\(\text{hoặc }3x-1=-5\)

        \(3x\)       \(=\left(-5\right)+1=-4\)

         \(x\)        \(=\left(-4\right):3=\dfrac{-4}{3}\)

\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)

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

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

\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)

       \(x\)         \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)

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

\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)

\(\left(4x-3\right)^2=16\)

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

\(\text{Vậy }4x-3=4\) 

        \(4x\)       \(=4+3=7\)

          \(x\)       \(=7:4=\dfrac{7}{4}\)

\(\text{hoặc }4x-3=-4\)

        \(4x\)        \(=\left(-4\right)+3=-1\)

          \(x\)        \(=\left(-1\right):4=\dfrac{-1}{4}\)

\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)

31 tháng 7 2023

\(4.3^x+3^{x+1}=63\)

\(\Rightarrow4.3^x+3.3^x=63\)

\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)

\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)

mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)

\(\Rightarrow x\in\varnothing\)