cái này có những biểu thức ko dùng được hđt hoặc đặt nhân tử chung =))) anh nghĩ đề em nên thay chữ và thành hoặc 

1, \(15x^2+10xy=5x\left(3x+2y\right)\)

2, \(24x-18y+30=6\left(4x-3y+5\right)\)

3, \(2x\left(y-2009\right)+5y\left(y-2009\right)=\left(y-2009\right)\left(2x+5y\right)\)

4, \(35x\left(y-8\right)-14y\left(8-y\right)=\left(y-8\right)\left(35x+14y\right)=7\left(5x+2y\right)\left(y-8\right)\)

5, \(x^2+14x+49=x^2+2.7x+7^2=\left(x+7\right)^2\)

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

Trả lời:

1, 15x² + 10xy = 5x ( 3x + 2y )

2, 24x - 18y + 30 = 6 ( 4x - 3y + 5 )

3, 2x ( y - 2009 ) + 5y ( y - 2009 ) = ( y - 2009 )( 2x + 5y )

4, 35x ( y - 8 ) - 14y ( 8 - y ) = 35x ( y - 8 ) + 14y ( y - 8 ) = ( y - 8 )( 35x + 14y ) = 7 ( y - 8 )( 5x + 2y )

5, x² + 14x + 49 = ( x + 7 )² 

6, 9x² - 4 = ( 3x - 2 )( 3x + 2 )

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

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

4) \(x^2-2xy-16z^2+y^2=\left(x-y\right)^2-\left(4z\right)^2\)

\(=\left(x-y-4z\right)\left(x-y+4z\right)\)

5) \(x^3+x^2y-x^2z-xyz=x^2\left(x-z\right)+xy\left(x-z\right)\)

\(=\left(x^2+xy\right)\left(x-z\right)\)

6) \(\text{Đề bài}=\left(3x\right)^2-\left(2x-2\right)^2=\left(3x-2x+2\right)\left(3x+2x-2\right)\)

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

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

\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(a+b+c\right)\)

8) \(x^3-8y^6=x^3-\left(2y^2\right)^3=\left(x-2y^2\right)\left(x^2+2xy^2+4x^2y^4\right)\)

9) \(x^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)

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

10) \(x^4+x^2y^2+y^4=x^4+2x^2y^2+y^2-x^2y^2\)

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

a) 85.12,7 + 5.3.12,7=12,7(85+15)=1270

b) 8,4.84,5 + 840.0,155;=8,4.84,5+8,4+15,5

=8,4(84,5+18400

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

2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)

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

4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)

5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)

\(=4x\left(x+y\right)-9y\left(x+y\right)=\left(4x-9y\right)\left(x+y\right)\)

6, \(x^2-2x-4y^2-4y=x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2\)

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

a, ĐK : \(x\ne\pm3;\frac{1}{2}\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)

\(=\left(\frac{\left(x-1\right)\left(x-3\right)+2\left(x+3\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x-1-2x-1}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)

\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)

b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)

TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)

Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)

TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)

Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)

c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)

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

b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)

\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

a) x\(^2\) - 10x + 9 =0

x\(^2\) - 2x . 5 + 25 = 16

(x - 5)\(^2\) = 4\(^2\)

=> x - 5 = 4

x = 9

Vậy x = 9

b) x\(^2\) - 7x + 6 = 0

x\(^2\) - 2x . 3,5 + 12,25 = 6,25

(x - 3,5)\(^2\) = 2,5\(^2\)

=> x - 3,5 = 2,5

x = 6

Vậy x = 6

c) x\(^2\) + 13x + 12 = 0

x\(^2\) + 2x . 6,5 + 42,25 = 30,25

(x + 6,5)\(^2\) = 5,5\(^2\)

=> x + 6,5 = 5,5

x = -1

Vậy x = -1

d) x\(^2\) - 24x + 23 = 0

x\(^2\) - 2x . 12 + 244 = 121

(x - 12)\(^2\) = 11\(^2\)

=> x - 12 = 11

x = 23

Vậy x = 23

e) 3x\(^2\) + 14x + 8 = 0

3x\(^2\) + 2 . \(\sqrt{3}\)x . \(\frac{7}{\sqrt{3}}\) + \(\frac{49}{3}\) = \(\frac{25}{3}\)

(\(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\))\(^2\) = \(\left(\frac{5}{\sqrt{3}}\right)^2\)

=> \(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\) = \(\frac{5}{\sqrt{3}}\)

=> \(\sqrt{3}\)x  = \(\frac{-2}{\sqrt{3}}\)

=> x = \(\frac{-2}{3}\)

Đặt a - 2b + c = x 

b - 2c + a = y

c - 2a + b = z

Ta có x + y + z = 0

=> x + y = - z

=> (x + y)³ = (-z)³

=> x³ + y³ + 3xy(x + y) = - z³ 

=> x³ + y³ + z³ = -3xy(x + y) 

=> x³ + y³ + z³ = 3xyz

mà x³ + y³ + z³ = 0 

=> 3xyz = 0 

=> xyz = 0

=> x = 0 hoặc y = 0 hoặc z = 0

Khi x = 0 => a - 2b + c = 0 

=> a + c = 2b

=> a + b + c = 3b \(⋮\)3

Khi y = 0 => b - 2c + a = 0

=> a + b = 2c 

=> a + b + c = 3c \(⋮\)3

Khi z = 0 => c - 2a + b = 0

=> b + c = 2a

=> a + b + c = 3a \(⋮\)3

Vậy a + b + c \(⋮3\)

\(a)\)

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

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

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

\(=x^4-1\)

\(b)\)

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

\(=x^5+1^5\)

\(=x^5+1\)