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

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

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

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=30:15\)

\(\Leftrightarrow x=2\)

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

<=> 36x² - 12x - 36x² + 27x = 30

<=> 15x = 30

<=> x = 2

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

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

\(\Leftrightarrow15x=30\)

\(\Rightarrow x=2\)

 3x(12x-4)-9x(4x-3)= 30

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

<=>15x = 30

=>x= 2

Bài giải:

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

36x² – 12x – 36x² + 27x = 30

15x = 30

Vậy x = 2.

b) x (5 - 2x) + 2x (x - 1) = 15

5x – 2x² + 2x² – 2x = 15

3x = 15

x =5

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

36x² – 12x – 36x² + 27x = 30

15x = 30

Vậy x = 2.

b) x (5 - 2x) + 2x (x - 1) = 15

5x – 2x² + 2x² – 2x = 15

3x = 15

x =5

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

\(\Leftrightarrow\) 36x² - 12x - 36x² + 27x = 30

\(\Rightarrow\) 15x = 30

\(\Rightarrow\) x = 2

b) x (5 - 2x) + 2x (x - 1) = 15

\(\Leftrightarrow\) 5x - 2x² + 2x² - 2x = 15

\(\Rightarrow\) 3x = 15

\(\Rightarrow\) x = 5

a) ta có : \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)

b) điều kiện : \(x\ne\dfrac{1}{5};x\ne1;x\ne\dfrac{3}{5}\)

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

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

\(\Leftrightarrow\dfrac{x+7}{3-3x}=\dfrac{4}{3-5x}\Leftrightarrow\left(x+7\right)\left(3-5x\right)=4\left(3-3x\right)\)

\(\Leftrightarrow-5x^2-20+9=0\)

ta có : \(\Delta'=\left(10\right)^2+5\left(9\right)=145>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x=\dfrac{10+\sqrt{145}}{-5};x=\dfrac{10-\sqrt{145}}{-5}\)

a) \(36x^2-12x-36x^2+27x=30\)

                                                \(15x=30\)

                                                     \(x=2\)

b) \(5x-2x^2+2x^2-2x=15\)

                                          \(3x=15\)

                                             \(x=5\)

ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

Vì trong bài làm của mình có một số dòng khá dài nên bạn có thể vào trang cá nhân của mình để đọc tốt hơn!

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

b: \(=\dfrac{2x^3-10x^2-17x^2+85x+30x-150}{x-5}=2x^2-17x+30\)

c: \(=\dfrac{12x^4-8x^3+12x^3-8x^2+8x^2-\dfrac{16}{3}x+\dfrac{43}{3}x-\dfrac{86}{9}+\dfrac{113}{9}}{3x-2}\)

\(=4x^3+4x^2+\dfrac{8}{3}x+\dfrac{43}{9}x+\dfrac{\dfrac{113}{9}}{3x-2}\)