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 \(x=\dfrac{2}{3}\)

a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

=(x-3)(x+2)(x+1)

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

=x^2(2x-3)-x(2x-3)+(2x-3)

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

d: =2x^3+x^2+2x^2+x+2x+1

=(2x+1)(x^2+x+1)

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

\(=\left(x^3-x\right)-\left(6x+6\right)\)

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

\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)

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

b) \(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

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

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

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

c) \(2x^3-5x^2+5x+1\)

\(=2x^3-3x^2-2x^2+3x+2x-3\)

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

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

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

d) \(2x^3+3x^2+3x+1\)

\(=2x^3+x^2+2x^2+x+2x+1\)

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

\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)

\(=\left(2x+1\right)\left(x^2+x+1\right)\)

e) \(3x^3-2x^2+5x+2\)

\(=3x^3+x^2-3x^2-x+6x+2\)

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

$\textbf{5)}$

$(4x-5)(x+2)-(x+5)(x-3)-3x^2-x$

$=(4x^2+3x-10)-(x^2+2x-15)-3x^2-x$

$=4x^2+3x-10-x^2-2x+15-3x^2-x$

$=5.$

a: =(6x)^2-(3x-2)^2

=(6x-3x+2)(6x+3x-2)

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

d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)

\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)

=8x(x^2+1)

e: =(4x)^2-2*4x*3y+(3y)^2

=(4x-3y)^2

f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)

\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)

g: =(4x)^3+1^3

=(4x+1)(16x^2-4x+1)

k: =x^3(27x^3-8)

=x^3(3x-2)(9x^2+6x+4)

l: =(x^3-y^3)(x^3+y^3)

=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)

$36x^2-(3x-2)^2$

$=[6x-(3x-2)][6x+(3x-2)]$

$=(6x-3x+2)(6x+3x-2)$

$=(3x+2)(9x-2)$

$16(4x+5)^5-25(2x+2)^2$

Câu này không có dạng hằng đẳng thức hiệu hai bình phương vì số mũ của hai phần là $5$ và $2$.

Có thể đặt nhân tử chung $1$ nhưng không phân tích tiếp được bằng các phương pháp thông thường.

$(x-y+4)^2$

Đây đã là một tích:

$(x-y+4)^2=(x-y+4)(x-y+4)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-x^2+2x-1][x^2+2x+1+x^2-2x+1]$

$=4x(2x^2+2)$

$=8x(x^2+1)$

$16x^2-24xy+9y^2$

$=(4x)^2-2\cdot4x\cdot3y+(3y)^2$

$=(4x-3y)^2$

$-\dfrac{x^4}{4}+2x^2y^3-4y^6$

$=-\dfrac14(x^4-8x^2y^3+16y^6)$

$=-\dfrac14(x^2-4y^3)^2$

$64x^3+1$

$=(4x)^3+1^3$

$=(4x+1)(16x^2-4x+1)$

$x^3y^6z^9-125$

$=(xy^2z^3)^3-5^3$

$=(xy^2z^3-5)(x^2y^4z^6+5xy^2z^3+25)$

$27x^6-8x^3$

$=x^3(27x^3-8)$

$=x^3[(3x)^3-2^3]$

$=x^3(3x-2)(9x^2+6x+4)$

$x^6-y^6$

$=(x^3-y^3)(x^3+y^3)$

$=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

$27x^3-54x^2y+36xy^2-8y^3$

$=(3x)^3-3(3x)^2(2y)+3(3x)(2y)^2-(2y)^3$

$=(3x-2y)^3$

$y^9-9x^2y^6+27x^4y^3-27x^6$

$=(y^3)^3-3(y^3)^2(3x^2)+3(y^3)(3x^2)^2-(3x^2)^3$

$=(y^3-3x^2)^3$

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

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

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

\(\Leftrightarrow x+3=0\)

hay x=-3

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

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

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

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

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

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

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

x(2x-4)-(x-2)(2x+3)

<=> (2x² - 4x ) - (2x² - 3x - 4x -6=0

<=> 2x² - 4x -2x² -3x - 4x  + 6 =0

<=> -3x + 6 =0

<=> -3x = 6 

<=> x = -(6/3) = -2

Bạn ơi bài mik đâu cs bằng 0 đâu

a: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>\(\left(x+3\right)\cdot\left(27x^2-12x\right)=0\)

=>3x(x+3)(9x-4)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

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

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

=>(x-2)(3x+5)-(x-2)(2x+2)=0

=>(x-2)(3x+5-2x-2)=0

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

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

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

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

=>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)