`1)x^3-7x+6`

`=x^3-x-6x+6`

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

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

`=(x-1)(x^2-2x+3x-6)`

`=(x-1)[x(x-2)+3(x-2)]`

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

`2)x^3-9x^2+6x+16`

`=x^3-2x^2-7x^2+14x-8x+16`

`=x^2(x-2)-7x(x-2)-8(x-2)`

`=(x-2)(x^2-7x-8)`

`=(x-2)(x^2-8x+x-8)`

`=(x-2)[x(x-8)+x-8]`

`=(x-2)(x-8)(x+1)`

`3)x^3-6x^2-x+30`

`=x^3+2x^2-8x^2-16x+15x+30`

`=x^2(x+2)-8x(x+2)+15(x+2)`

`=(x+2)(x^2-8x+15)`

`=(x+2)(x^2-3x-5x+15)`

`=(x+2)[x(x-3)-5(x-3)]`

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

`4)2x^3-x^2+5x+3`

`=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)`

`5)27x^3-27x^2+18x-4`

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

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

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

1) Ta có: \(x^3-7x+6\)

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

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

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

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

2) Ta có: \(x^3-9x^2+6x+16\)

\(=x^3-2x^2-7x^2+14x-8x+16\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)\)

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

3) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

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

Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

cảm ơn nhé

a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)

b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)

c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)

d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)

a) 

/ \(x^4+x^2-2=0\)

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

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

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

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

\(=x^2+4xy+4y^2\)

e) \(\left(x-3y\right)\left(x+3y\right)\)

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

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

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

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

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

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

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

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

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

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

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

(x - 1)(x - 2)(x - 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3

a) x³y + x - y - 1

= (x³y - y) + (x - 1)

= y(x³ - 1) + (x - 1)

= y(x - 1)(x² + x + 1) + (x - 1)

= (x - 1)[y(x² + x + 1) + 1]

= (x - 1)(x²y + xy + y + 1)

b) x²(x - 2) + 4(2 - x)

= x²(x - 2) - 4(x - 2)

= (x - 2)(x² - 4)

= (x - 2)(x - 2)(x + 2)

= (x - 2)²(x + 2)

c) x³ - x² - 20x

= x(x² - x - 20)

= x(x² + 4x - 5x - 20)

= x[(x² + 4x) - (5x + 20)]

= x[x(x + 4) - 5(x + 4)]

= x(x + 4)(x - 5)

d) (x² + 1)² - (x + 1)²

= (x² + 1 - x - 1)(x² + 1 + x + 1)

= (x² - x)(x² + x + 2)

= x(x - 1)(x² + x + 2)

e) 6x² - 7x + 2

= 6x² - 3x - 4x + 2

= (6x² - 3x) - (4x - 2)

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

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

f) x⁴ + 8x² + 12

= x⁴ + 2x² + 6x² + 12

= (x⁴ + 2x²) + (6x² + 12)

= x²(x² + 2) + 6(x² + 2)

= (x² + 2)(x² + 6)

g) (x³ + x + 1)(x³ + x) - 2

Đặt u = x³ + x

x³ + x + 1 = u + 1

(u + 1).u - 2

= u² + u - 2

= u² - u + 2u - 2

= (u² - u) + (2u - 2)

= u(u - 1) + 2(u - 1)

= (u - 1)(u + 2)

= (x³ + x - 1)(x³ + x + 2)

= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)

= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]

= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]

= (x³ + x - 1)(x - 1)(x² - x + 2)

h) (x + 1)(x + 2)(x + 3)(x + 4) - 1

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1

= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)

Đặt u = x² + 5x + 4

u + 2 = x² + 5x + 6

(1) u.(u + 2) - 1

= u² + 2u - 1

= u² + 2u + 1 - 2

= (u² + 2u + 1) - 2

= (u + 1)² - 2

= (u + 1 + √2)(u + 1 - √2)

= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)

= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)

\(c,20=2^2\cdot5\\ 45=3^2\cdot5\\ ƯCLN\left(20,45\right)=5\\ \RightarrowƯC\left(20,45\right)=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\\ C=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(d,\left(6x^2-7x+1\right)\left(x^3-x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-1\right)x\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)

Sửa: \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-x-1\right)\)

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

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

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

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

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)