\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

<=> \(x^3-9x^2+27x-27\) \(-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)+3x^2=-33\)

<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)

<=> \(-6x^2+39x+6=-33\)

<=> \(6x^2-39x-6=33\)

<=> \(6x^2-39x-39=0\)

<=> \(6\left(x^2-\frac{39}{6}x-\frac{39}{6}\right)=0\)

<=> \(x^2-2.x.\frac{39}{12}+\frac{1521}{144}-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}\right)^2-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}-\frac{\sqrt{273}}{4}\right)\left(x-\frac{39}{12}+\frac{\sqrt{273}}{4}\right)=0\)

<=> \(\left(x-\frac{13+\sqrt{273}}{4}\right).\left(x-\frac{13-\sqrt{273}}{4}\right)=0\)

<=> \(x=\frac{13+\sqrt{273}}{4}\) ( h ) \(x=\frac{13-\sqrt{273}}{4}\)

học tốt

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

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

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

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

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

\(\Leftrightarrow7x=1\)

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

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

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

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

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

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

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

`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):}`

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

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

Rảnh rỗi thật sự .-.

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

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

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

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

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

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

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

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

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

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

a: \(y'< 0\)

=>\(\left(x-3\right)^3\cdot\left(x-1\right)^{22}\cdot\left(-3x-6\right)^7< 0\)

=>\(\left(x-3\right)\left(-3x-6\right)< 0\)

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

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

y'>0

=>\(\left(x+2\right)\left(x-3\right)< 0\)

=>\(-2< x< 3\)

y'=0

=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\\-3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)

Ta có bảng xét dấu sau:

Vậy: Hàm số đồng biến trên các khoảng \(\left(-2;1\right);\left(1;3\right)\)

Hàm số nghịch biến trên các khoảng \(\left(-\infty;-2\right);\left(3;+\infty\right)\)

b: y'<0

=>\(\left(4x-3\right)^3\cdot\left(x^2-1\right)^{21}\left(3x-9\right)^7< 0\)

=>\(\left(4x-3\right)\left(3x-9\right)\left(x^2-1\right)< 0\)

=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)< 0\)

TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{3}{4}\)

TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)

y'>0

=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)>0\)

TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\-1< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{3}{4}< x< 1\)

Ta sẽ có bảng xét dấu sau đây:

Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;-1\right);\left(\dfrac{3}{4};1\right);\left(3;+\infty\right)\)

Hàm số nghịch biến trên các khoảng \(\left(-1;\dfrac{3}{4}\right);\left(1;3\right)\)