cac ban oi truoc cho 3x co dau ngoac va sau so 10 cung co dau ngoac

3x - 10: 10  = 50

=>. 3x -1 = 50

=> 3x = 50 +1

=> 3x = 51

=> x = 51: 3

=> x = 17

3x-10:10=50

      3x-1=50

         3x=50+1

         3x=51

           x=51:3

           x=17

\(\dfrac{3x-40}{50}+\dfrac{3x-10+2x+60+4x-360}{40}=0\)

=> \(\dfrac{3x-40}{50}+\dfrac{9x-310}{40}=0\)

=> \(\dfrac{3x-40}{50}=\dfrac{-9x+310}{40}\)

=> \(40\left(3x-40\right)=50\left(-9x+310\right)\)

=> \(120x-1600=-450x+15500\)

=> \(120x+450x=15500+1600\)

Hay \(570x=17100\)

=>x = 30

Hơi dài nhé bạn

\(\dfrac{3x-40}{50}\)+\(\dfrac{3x-10+2x+60+4x-360}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)+\(\dfrac{9x-310}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)=\(\dfrac{9x-310}{40}\)

⇒40(3x -40) = 50(-9x+310)

⇒120x - 1600 = -450x + 15500

⇒120x + 450x = 15500 + 1600

Mặt khác: 570x = 17100

⇒x = 30

LÀM LUN NHA

2/3x-1/2+x=1/10

5/3x=1/10+1/2

5/3x=3/5

X=9/25

**** CHO MK NHA

a) Ta có: \(\left|x\right|-\left(-9\right)^3=-100\)

\(\Leftrightarrow\left|x\right|-\left(-729\right)=-100\)

\(\Leftrightarrow\left|x\right|+729=-100\)

\(\Leftrightarrow\left|x\right|=-100-729=-829\)(vô lý)

Vậy: \(x\in\varnothing\)

b) Ta có: \(3x-\left(-10\right)=2x+50\)

\(\Leftrightarrow3x+10-2x-50=0\)

\(\Leftrightarrow x-40=0\)

\(\Leftrightarrow x=40\)

Vậy: x=40

\(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)

\(\Leftrightarrow-62x+12=-50\)

\(\Leftrightarrow-62x=-62\)

\(\Leftrightarrow x=1\)

Có (8x + 2)(1 - 3x) + (6x - 1)(4x - 10) = -50

=>8x - 24\(^{x^2}\) + 2 - 6x + 24\(^{x^2}\) - 60x - 4x + 10 = -50

=> -62x + 12 = -50

=> -62x = -62

=> x=1 (T/m)

Vậy x=a

a)

   \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

   \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)

    \(\frac{1}{3}:x=\frac{-1}{15}\)

     \(x=\frac{1}{3}:\frac{-1}{15}\)

    \(x=-5\)

b)

     \(\frac{10}{3x}+\frac{67}{4}=-13.25\)

      \(\frac{10}{3x}+\frac{67}{4}=\frac{-53}{4}\)

       \(\frac{10}{3x}=\frac{-53}{4}+\frac{67}{4}\)

       \(\frac{10}{3x}=\frac{7}{2}\)

      \(\Rightarrow10.2=7.3x\)

            \(20=21x\)

            \(x=\frac{20}{21}\)

c)

        x+30%x=-1.3

        x+0.3x=-1.3

        x (1+0.3) = -1.3

        x . 1.3 = -1.3

        x = -1.3 : 1.3

        x = 1

d)

          \(\left(\frac{14}{5x}-50\right):\frac{2}{3}=51\)

           \(\frac{14}{5x}-50=51.\frac{2}{3}\)

             \(\frac{14}{5x}-50=34\)

             \(\frac{14}{5x}=34+50\)

              \(\frac{14}{5x}=84\)

               \(84.5x=14\)

              420x = 14

                   \(x=\frac{1}{30}\)

a, \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

=> \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)

=>\(\frac{1}{3}:x=\frac{-1}{15}\)

=> \(x=\frac{1}{3}:\frac{-1}{15}\)

=> \(x=\frac{-1}{5}\)

mình sai chỗ kết quả cuối bằng -5 nhé

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

c)

\(4\left(3x-4\right)-2=18\)

<=> \(12x-16-2=18\)
<=> \(12x=36\)

<=> \(x=3\)

Vậy x=3

d)

\(\left(3x-10\right):10=50\)

<=> \(3x-10=500\)

<=> \(3x=510\)

<=> x= \(170\)

Vậy x= 170

f)

\(x-\left[42+\left(-25\right)\right]=-8\)

<=> \(x-17=-8\)

<=> x= \(9\)

Vậy x=9

h)

\(x+5=20-\left(12-7\right)\)

<=> \(x+5=15\)

<=> \(x=10\)

Vậy x= 10

k)

\(\left|x-5\right|=7-\left(-3\right)\)

<=> \(\left|x-5\right|=10\)

* Với \(x>=5\) ; ta được:

\(x-5=10\)

<=> x= 15 (thoả mãn điều kiện )

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=10\)

<=> \(-x+5=10\)

<=> \(-x=5\)

<=> \(x=-5\) (thoả mãn điều kiện)

Vậy x=15 ; x= -5

i)

\(\left|x-5\right|=\left|7\right|\)

<=> \(\left|x-5\right|=7\)

*Với \(x>=5\) ; ta được:

\(x-5=7\)

<=> \(x=12\) (thoả mãn)

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=7\)

<=> \(-x=2\)

<=> \(x=-2\) (thoả mãn)

Vậy x= 12; x= -2

m)

\(2^{x+1}.2^{2009}=2^{2010}\)

<=> \(2^{x+1+2009}=2^{2010}\)

<=> \(2^{x+2010}=2^{2010}\)

=> \(x+2010=2010\)

=> \(x=0\)

Vậy x=0

n)

\(10-2x=25-3x\)

<=>\(x=15\)

Vậy x=15

nik lộn

2^x+1. 2^{2009 =} 2²⁰¹⁰

10 - 2x = 25 - 3x