a)128-3(x+4)=23 

    125.(x+4)=23

             ( x +4)=23 : 125

             x+4 = 0,184

                x = 0,184 -4

                  x = -3 .816

theo mềnh nhóe ( làm từng câu )

c: =>12x-64=32

=>12x=96

hay x=8

g: =>3x-16=14

hay x=10

e: =>5(x+4)=85

=>x+4=17

hay x=13

a) (x + 3)^2 - 4x - 12 = 0

<=> (x + 3)^2 - 4(x + 3) = 0

<=> (x + 3)(x - 1) = 0

<=> x = -3 hoặc x = 1

b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7

<=> x^3 - 25x - x^3 + 27 = 7

<=> -25x + 27 = 7

<=> x = 4/5

a/ \(\left(x+3\right)^2-4x-12=0\)

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

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

\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)

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b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)

\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)

\(x^3-25x-x^3+27=7\)

\(-25x=-20\)

\(x=\dfrac{20}{25}=\dfrac{4}{5}\)

a, <=>x² +6x+9-4x-12=0

<=> x² +2x -3=0

<=> x² +3x -x-3=0

<=> x.(x+3) - (x+3) =0

<=> (x-1)(x+3)=0

<=> x=1 hoặc x=-3

b, <=> x(x² -25) - (x-3)(x+3)² -7=0 

<=> x³ -25x + (9-x²) (x+3) -7=0

<=> x³ -25x+ 9x+27-x³ -3x² -7=0

<=> -3x² -16x +20=0

<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)

<=> x= 10/3 hoặc x=2

Chúc bạn học tốt nha!

a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1

a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

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

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)

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

\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

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

a: \(x\cdot3\cdot5=2,7\)

=>\(x\cdot15=2,7\)

=>\(x=\dfrac{2,7}{15}=0,18\)

b: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)