\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-3^3\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9.\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\Leftrightarrow x=\frac{2}{15}\)

Vậy: \(S=\left\{\frac{2}{15}\right\}\)

pt <=> \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)

   <=> \(x^3-3x^2.3+3x.3^2-27-x^3+27+9x^2+18x+9=15\)

   <=>  \(45x=6\)

   <=>  \(x=\frac{6}{45}=\frac{2}{15}\)

= \(\frac{2}{15}\) nha bn

a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

=>4x-27=1

hay x=7

b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)

\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)

=>39x+6=15

hay x=3/13

c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)

\(\Leftrightarrow3x-40=2\)

hay x=14

a. x = 9

b. x = 5

c. x = 8

Đề nhìn vô lí quá

a. x = 9

b. x = 5

c. x = 8

Tham hkaor

a. x = 9

b. x = 5

c. x = 8

Đề nhìn vô lí quá

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

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

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

<=>(x³-9x²+27x-27)-(x³-3³+6(x²+2x+1)=15

<=>x³-9x²+27x-27-x³+27+6x²+12x+6=15

<=>-3x²+39x+9=0

<=>x²-13x+3=0

<=>(x²-2.x.13/2+169/4)-157/4=0

<=>(x-13/2)²=157/4

<=>x-13/2=\(\sqrt{\frac{157}{2}}\)hoặc x=13/2= - \(\sqrt{\frac{157}{2}}\)

<=>x=(13+\(\sqrt{\frac{157}{2}}\))hoặc x=\(\frac{13-\sqrt{\frac{157}{2}}}{2}\)

(x-3)³ - (x-3)(x²+3x+9) + 6(x+1)² = 15

 x³ -9x² + 27x - 27 - (x³-27) + 6( x²+ 2x + 1)  =15

 x³ -9x² + 27x - 27 - x³+ 27 + 6x² + 12x + 6 = 15

-3x² + 39x -9 = 0

-3(x² - 13x + 3) = 0

x² - 13x + 3 = 0

=> x=0,2350179569 ( chỗ này bấm máy tính)

còn giải thì làm theo mấy cách trong đây 

BÀI 3 – 4

Phương trình bậc hai một ẩn – Công thức nghiệm

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

Sửa đề: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2-3x+9\right)+9\left(x+1\right)^2=15\)

Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2-3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3-27\right)+9\left(x^2+2x+1\right)-15=0\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27+9x^2+18x+9-15=0\)

\(\Leftrightarrow6x^2+21x+20=0\)

\(\Leftrightarrow6\left(x^2+\frac{7}{2}x+\frac{10}{3}\right)=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\frac{7}{4}+\frac{49}{16}+\frac{13}{48}=0\)

\(\Leftrightarrow\left(x+\frac{7}{4}\right)^2+\frac{13}{48}=0\)(Vô lý)

Vậy: x∈∅

(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2

=>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x-40=2

=>x=42/3=14

giúp mik mik đang cần gấp

nhưng phả có lời giải đừng cho mỗi đáp án

a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)

\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)

b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)

\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)

bài này hôm qua giải rồi mà!!