Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

\(a,=10x-8y+3\\ b,=\left(x+y\right)^2:\left(x+y\right)=x+y\\ c,=\left(a+b\right)\left(3a-2c+d\right):\left(a+b\right)=3a-2c+d\\ d,=\left(x-1\right)^3:\left(x-1\right)^2=x-1\)

\(x^2-2xy+y^2-a^2+4ab-4b^2\)

\(=\left(x-y\right)^2-\left(a-2b\right)^2\)

\(=\left(x-y-a+2b\right)\left(x-y+a-2b\right)\)

hk tốt

^^

Sửa đề: B=-2x^2+xy+2y^2-3-5x+2y

a: A+B+C

=x^2-3xy-y^2+2x-3y+1-2x^2+xy+2y^2-3-5x+2y+C

=-x^2-2xy+y^2-3x-y-2+3x^2+7y^2-4xy-6x+4y+5

=2x^2+8y^2-6xy-9x+3y+3

b: 7A-B-C-9

=7A-9-(x^2+9y^2-3xy-11x+6y+2)

=7x^2-7y^2-21xy+14x-21y+7-x^2-9y^2+3xy-11x-6y-2-9

=6x^2-16y^2-18xy+3x-27y-4

Bài 1 : 

a) \(x^2+y^2\)

\(\Leftrightarrow x^2+2xy+y^2-2xy\)

\(\Leftrightarrow\left(x+y\right)^2-2xy=\left(-3\right)^2-2.\left(-28\right)=65\)

b) \(x^3+y^3\)

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

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

\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(-3\right)\left[\left(-3\right)^2-3.\left(-28\right)\right]=-279\)

c) \(x^4+y^4\)

\(\Leftrightarrow\left(x+y\right)^4-4x^3y-4xy^3-6x^2y^2=\left(-3\right)^4-4\left(-28\right).65-6\left(-28\right)^2=2657\)

Bài 3:

Có:    \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

=>     \(x^3+y^3+z^3=\left(-z\right)^3-3xy.-z+z^3\)

=>     \(x^3+y^3+z^3=-z^3+z^3+3xyz=3xyz\)

=> TA CÓ ĐPCM.

VẬY      \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Bài 2 :

a) Giả sử  \(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-ab-ac-ad\ge0\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2-4ab-4ac-4ad\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d^2\right)\ge0\)( luôn đúng )

\(\RightarrowĐPCM\)

b) Sửa đề : \(a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Ta có : \(\left(a+2c\right)^2\ge0\Leftrightarrow a^2+4c^2\ge-4ac\left(1\right)\)

Áp dụng BĐT Cô - si ta có :

\(\hept{\begin{cases}a^2+4b^2\ge4ab\left(2\right)\\4b^2+4c^2\ge8bc\left(3\right)\end{cases}}\)

(1) + (2) + (3) 

\(\Leftrightarrow2a^2+8b^2+8c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow2\left(a^2+4b^2+4c^2\right)\ge4\left(ab-ac+2bc\right)\)

\(\Leftrightarrow a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Bài 1 :

a) (3a+4b)³+(3a-4b)³-48a²b²

=27a³+108a²b+144ab²+64b³+27a³-108a²b+144ab²-64b³-48a²b²

=54a³+288ab²-48a²b²

=2a(27a²+144b²-24ab)

b) (5x+2y)(5x-2y)+(2x-y)³+(2x+y)³

=25x²-4y²+8x³-12x²y+6xy²-y³+8x³+12x²y+6xy²+y³

=16x³+25x²-y²+12xy²

=x²(16x+25)-y²(1-12x)

Bài 2 :

\(x^2-8x+7=0\)

\(\Leftrightarrow x^2-x-7x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)

c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn

d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)

\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)

\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)

\(\Leftrightarrow27x^3-54x^2-27x=0\)

\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

#H

năm nay tuấn 8 tuổi , chị hoa hơn tuấn 5 tuổi . tuổi mẹ gấp 3 lần tuổi chhj hoa. hỏi năm nay mẹ bao nhiêu tuổi ?

1

a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)

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

\(=6x^2+5x+1-6x^2-6x+x+1\)

\(=2\)

c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)

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

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

\(=a^3+a^2\)

2,

a,\(4ab+a^2-3a-12b\)

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

\(=4b\left(a-3\right)+a\left(a-3\right)\)

\(=\left(4b+a\right)\left(a-3\right)\)

b,\(x^3+3x^2+3x+1-27y^3\)

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

\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)

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

4

a,\(2004^2-16\)

\(=2004^2-4^2\)

\(=\left(2004-4\right)\left(2004+4\right)\)

\(=2000.2008\)

\(=4016000\)

b,\(892^2+892.216+108^2\)

\(=\left(892+108\right)^2\)

\(=1000^2=1000000\)

c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)

\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)

\(=9,8.10+10,2.10\)

\(=98+102\)

\(=200\)

d,\(36^2+26^2-52.36\)

=\(\left(36-26\right)^2\)

\(=10^2=100\)

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

\(\Leftrightarrow A=-x^2+2x-1-2\)

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

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

Vậy GTLN của A=-2 khi x=1

a) x³ - 2x² - 9x + 18

= x²( x - 2) - 9( x - 2)

= ( x² - 3²).( x -2)

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

b) x² - a² - 4ab - 4b²

= -[ a² + 2.2ab + (2b)² - x²]

= -[ ( a + 2b)² - x²]

= -( a + 2b - x).( a +2b +x)

c) x² - x - 2

= x² - 2x + x -2

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

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

d) x² + x -6

= x² - 2² + x -2

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

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

e) x³ - 3x² - 4x + 12

= x²( x - 3) - 4( x - 3)

= ( x² - 2²).( x -3)

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

Câu 1: 

a: \(=6x^2+5x+1-6x^2-6x+x+1=2\)

b: \(=a^3+1-a^2+1=a^3-a^2+2\)