\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

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

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

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

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

a: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

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

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

c: \(x^2-2xy+y^2-5x+5y\)

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

\(=\left(x-y\right)\left(x-y-5\right)\)

a) 

b) 

c) 

d) 

e) 

f) 

g) h) 

i) 

a.

\(x^3-7x+6=0\)

\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)

f.

\(x^4-4x^3+12x-9=0\)

\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

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

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

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

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

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

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

g.

\(x^5-5x^3+4x=0\)

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

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

\(\Leftrightarrow x\left[x^2\left(x^2-1\right)-4\left(x^2-1\right)\right]=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\end{matrix}\right.\)

1.

Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)

Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)

\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)

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

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

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

Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):

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

\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)

Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\): 

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)

Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

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

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

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

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

a) (5x³ – 2x² + 4x – 4) . ( x³ + 3x² – 5)

= 5x³ . ( x³ + 3x² – 5) - 2x² . ( x³ + 3x² – 5) + 4x . ( x³ + 3x² – 5) – 4 . ( x³ + 3x² – 5)

= 5x³ . x³ + 5x³ . 3x² + 5x³ . (-5) – [ 2x² . x³ + 2x² . 3x² +2x² . (-5)] + [4x . x³ + 4x. 3x² + 4x . (-5)] – [ 4x³ + 4.3x² + 4.(-5)]

= 5x⁶ + 15x⁵ – 25x³ – (2x⁵ + 6x⁴ – 10x²) + 4x⁴ + 12x³ – 20x – (4x³ + 12x² – 20)

= 5x⁶ + 15x⁵ – 25x³ – 2x⁵ - 6x⁴ + 10x² + 4x⁴ + 12x³ – 20x – 4x³ - 12x² + 20

= 5x⁶ + (15x⁵ – 2x⁵ ) + (- 6x⁴ + 4x⁴ ) + (-25x³ + 12x³ – 4x³ ) + (10x² - 12x² ) – 20x + 20

= 5x⁶ + 13x⁵ – 2x⁴ – 17x³ -2x² – 20x + 20

b) (-2,5.x⁴ + 0,5x² + 1) . (4x³ – 2x + 6)

= -2,5.x⁴ . (4x³ – 2x + 6) + 0,5x² . (4x³ – 2x + 6) + 1. (4x³ – 2x + 6)

= (-2,5.x⁴) . 4x³ + (-2,5.x⁴ ) . (-2x) + (-2,5.x⁴ ) . 6 + 0,5x² . 4x³ + 0,5x² . (-2x) + 0,5x² . 6 + 4x³ – 2x + 6

= -10x⁷ + 5x⁵ – 15x⁴ + 2x⁵ – x³ + 3x² + 4x³ – 2x + 6

= -10x⁷ + ( 5x⁵ + 2x⁵ ) - 15x⁴ + (– x³ + 4x³ ) + 3x² – 2x + 6

= -10x⁷ +7x⁵ - 15x⁴ + 3x³ + 3x² – 2x + 6