:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

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

2:

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

b: \(2\left(x-1\right)+x^2-x\)

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

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

c: \(3x^2+14x-5\)

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

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

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

=>\(2x^2-2x-2x^2=4\)

=>-2x=4

=>x=-2

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

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

=>\(x^2-3x-x^2-x+2=5\)

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

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

=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)

=>4x(2x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Bài 2:

a) \(x^2+y^2-9-2xy\)

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

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

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

b) \(4x^2-5x-9\)

\(=4x^2+4x-9x-9\)

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

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

\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)

\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)

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

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

\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)

\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)

\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)

Bài 2:

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

Đặt \(\left(x-2\right)=y\)thay vào (1) ta được:

\(\left(y-1\right)^2-\left(y+3\right)y=5\)

\(\Leftrightarrow y^2-2y+1-\left(y^2+3y\right)-5=0\)

\(\Leftrightarrow y^2-2y+1-y^2-3y-5=0\)

\(\Leftrightarrow-5y-4=0\)

\(\Leftrightarrow-5y=4\)

\(\Leftrightarrow y=\frac{-4}{5}\left(2\right)\)

Thay x-2 =y vào (2) ta được 

\(\Leftrightarrow x-2=\frac{-4}{5}\)

\(\Leftrightarrow x=\frac{6}{5}\)

Vậy ...

b) \(3x^2+5x-8=0\)

\(\Leftrightarrow3x^2-3x+8x-8=0\)

\(\Leftrightarrow3x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-8}{3}\end{cases}}}\)

Vậy ...

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

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

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

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

2) \(2x\left(x-2\right)-\left(x-2\right)^2\)

\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)

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

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

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

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

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

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

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

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

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

\(=\left(x+2\right)\left(-2x-10\right)\)

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

Phân tích đa thức thành nhân tử

27y²-9(x+y)²=\(9\left(3y^2-\left(x+y\right)^2\right)\)

=\(9\left(\sqrt{3}y+x+y\right)\left(\sqrt{3}y-x-y\right)\)

Rút gọn biểu thức

(2x⁴-x³+3x²): (-1/3x)

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

Câu 1: C

Câu 2: =x(x-2)*(x+2)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)