`@` `\text {Ans}`

`\downarrow`

`a)`

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

`=> 3x (12x-4) - 3*3x (4x - 3) = 30`

`=> 3x [12x - 4 - 3(4x-3)] = 30`

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

`=> 3x (-4+9)=30`

`=> 3x*5=30`

`=> 3x=6`

`=> x=2`

Vậy, `x=2`

`b)`

`x( 5 - 2x) + 2x( x - 1)`

`=> x(5-2x) + 2x^2 - 2x=15`

`=> 5x - 2x^2 + 2x^2 - 2x =15`

`=> 3x = 15`

`=> x=5`

Vậy, `x=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

b)  ta có  \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)

        \(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)

    \(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)

 \(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)

   \(\left(4x-5\right)\left(8x^2-70x\right)=0\)

=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)

\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\)     Vậy   \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc   \(x=\frac{35}{4}\)

a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)

\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)

\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)

\(\Leftrightarrow132x^2+216x=0\)

\(\Leftrightarrow12x\left(11x+18\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)

19 22 25 28 5(3x + 2) – 4(2x +3) x*(1 + 2x) 4(1 + x) – 3(2x-5) 4x–8(6) - X) 23/ ... 2x” – 4x + 3x – 6 = 2x” – X-6 (b) (x-3) = (x-3)(x-3) (c) (2x+y)(2x–y) = x* = x* – 3x ... (x - 6)” 7 (3x + 5)(x-6) 8 (8x + 2)(3x + 4) (4x – 1)(2x – 3) 10 (2x +5)* 11 (8x – 3)(2x + ... 27 (4x + 3y)(x + y) 28 (2x + 5)(5x – 2) (4x – 3y)(4x + y) 30 (7x + 2y)(3x + 4y) 24/ ...

\(a)=3x\cdot\left(2x-7-4x+5\right)=3x\cdot\left(-2x-2\right)=3x\cdot\left[-2\cdot\left(x+1\right)\right]\)

Rảnh rỗi thật sự .-.

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

\(5\left(2x-1\right)+4\left(8-3x\right)=-5\)

\(10x-5+32-12x=-5\)

\(-2x=5-5+32\)

\(x=\frac{32}{-2}\)

\(x=-16\)

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

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

\(15x=30\)

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

x = 2

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

$\Leftrightarrow -2(-x^2+3x-4)=0$

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.