3x(12x – 4) – 9x(4x – 3) = 30

3x.12x – 3x.4 – (9x.4x – 9x.3) = 30

36x2 – 12x – 36x2 + 27x = 30

(36x2 – 36x2) + (27x – 12x) = 30

15x = 30

x = 2

Vậy x = 2.

3x (12x -4 )- 9x (4x-3)= 30 

<=>36x²-12x-36x³+27x=30

<=>15x=30

<=>x=2

\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(36x^2-12x-36x^2+27x=30\)

\(15x=30\)

\(x=\frac{30}{15}\)

\(x=2\)

= 36X-12X-36X-27X=30

=36X-36X-12X-27X=30

=0-12X-27X

=-12X-27X=30

=X.(-12-27)=30

hình như mình tính sai

nếu công thức đúng thì k nha

\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=2\)

3x(12x+4)-9x(4x-3)=30

=>36.x²+12x-36.x²-27x=30

=>12x-27x=30

=>-15x=30

=>x=30:(-15)

=>x=-2

vậy......

3x.12x+3x.(-4)-9x.4x-9x.(-3)=30

36x²-12x-36x²-27x=30   

24x-9x=30

15x=30

=>x=30:15

=>x=2

3x(12x-4)-9x(4x-3)=30

3x12x+3x-4-9x4x-9x-3=30

36x^2-12x-36x^2+27x=30

(36^2-36x^2)+(-12x+27x)=30

       0         +     15x     =30

=> 15x=30

Vậy: x=30:15=2

36x²-12x-36x²+27x=30

15x=30

x=2

 3x(12x - 4 ) -9x (4x -3 ) = 30 
<=> 36x² - 12x - 36x²+27x = 30 
<=> 15x = 30 
<=> x=2

\(3x\left(12x-4\right)-9.\left(4x-3\right)=30\)

\(=>36x^2-12x-36x^2+27x=30\)

\(=>15x=30\)

\(x=30:15=2\)

Vậy x = 2.

~ Hok tốt ~

3x(12x-4)-9x(4x-3)=30

<=>36x²-12x-36x²+27x=30

<=>15x=30

=>x=30:15

=>x=2

chuyển vế sang r phân tích thành nhân tử, có thể dùng máy tính bỏ túi nhé bạn

câu 1: 9\(x^2\) + 12\(x\) + 5  =11

           (3\(x\))² + 2.3.\(x\) .2 + 2² + 1 = 11

           (3\(x\) + 2)²      =  11 - 1

             (3\(x\) + 2)²    = 10

               \(\left[{}\begin{matrix}3x+2=\sqrt{10}\\3x+2=-\sqrt{10}\end{matrix}\right.\)

                \(\left[{}\begin{matrix}3x=\sqrt{10}-2\\3x=-\sqrt{10}-2\end{matrix}\right.\)

                  \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{3}\\x=\dfrac{-\sqrt{10}-2}{3}\end{matrix}\right.\)

                 Vậy S = {\(\dfrac{-\sqrt{10}-2}{3}\); \(\dfrac{\sqrt{10}-2}{3}\)} 

  Câu 2: 6\(x^2\) + 16\(x\) + 12 = 2\(x^2\)

              6\(x^2\) + 16\(x\) + 12 - 2\(x^2\) = 0

              4\(x^2\) + 16\(x\) + 12 = 0

              (2\(x\))² + 2.2.\(x\).4 + 16 - 4 = 0

               (2\(x\) + 4)²   = 4

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

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

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

              S = { -3; -1}

3, 16\(x^2\) + 22\(x\) + 11 = 6\(x\) + 5

    16\(x^2\) + 22\(x\) - 6\(x\)  + 11 - 5 = 0

     16\(x^2\) + 16\(x\) + 6 = 0

      (4\(x\))² + 2.4.\(x\) . 2 + 2² + 2 = 0

       (4\(x\) + 2)² + 2 = 0 (1) 

Vì (4\(x\)+ 2)² ≥ 0 ∀ ⇒ (4\(x\) + 2)² + 2 > 0 ∀ \(x\) vậy (1) Vô nghiệm

             S = \(\varnothing\)

Câu 4. 12\(x^2\) + 20\(x\) + 10 = 3\(x^2\) - 4\(x\) 

            12\(x^2\) + 20\(x\) + 10 - 3\(x^2\) + 4\(x\) = 0

            9\(x^2\) + 24\(x\) + 10 = 0

           (3\(x\))² + 2.3.\(x\).4 + 16 - 6 = 0

          (3\(x\) + 4)² = 6

            \(\left[{}\begin{matrix}3x+4=\sqrt{6}\\3x+4=-\sqrt{6}\end{matrix}\right.\)

              \(\left[{}\begin{matrix}3x=-4+\sqrt{6}\\3x=-4-\sqrt{6}\end{matrix}\right.\)

              \(\left[{}\begin{matrix}x=\dfrac{\sqrt{6}-4}{3}\\x=-\dfrac{\sqrt{6}+4}{3}\end{matrix}\right.\)

                    S = {\(\dfrac{-\sqrt{6}-4}{3}\); \(\dfrac{\sqrt{6}-4}{3}\)}

a) \(3x\cdot\left(12x-4\right)-9x\cdot\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=\dfrac{30}{15}\)

\(\Leftrightarrow x=2\)

b) \(\left(2x+1\right)-5\left(x-2\right)=10\)

\(\Leftrightarrow2x+1-5x+10=10\)

\(\Leftrightarrow-3x+11=10\)

\(\Leftrightarrow-3x=10-11\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)