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a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
a) 2x2.(5x3-4x2y-7xy +1) =10x5-8x4y-14x3y+2x2 b) (5x -2y)(x2 -xy +1) =5x3-5x2y+5x-2x2y+2xy2-2y =5x3-7x2y+2xy2+5x-2y c) (\(\dfrac{1}{2}\)x -1)(2x -3) =x2-\(\dfrac{3}{2}\)x-2x+3 =x2-\(\dfrac{7}{2}\)x+3 d) (x +3y)2 =x2+6xy+9y2 e) (3x -2y)2 =9x2-12xy+4y2 g) (\(\dfrac{1}{4}\)x - 3y)(\(\dfrac{1}{4}\)x +3y) =\(\dfrac{1}{16}\)x2-9y2 f) (2x +3)3 =8x3+36x2+54x+27 h) (3 -2y)3 =27-54y+36y2-8y3
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
c) \(2x^3-x^2-8x+4\)
\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)
d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)
\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2+x\right)\)
e) \(2x^2-5x+2\)
\(=\left(2x^2-x\right)-\left(4x-2\right)\)
\(=x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
- Viết 7 hằng đẳng thức đáng nhớ :
\(\left(A+B\right)^2=A^2+2AB+B^2\)
\(\left(A-B\right)^2=A^2-2AB+B^2\)
\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)
\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)
\(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\)
\(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)
\(A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)\)
- Áp dụng :
\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)
\(b,\left(\dfrac{5x-1}{2}\right)^2=\dfrac{\left(5x-1\right)^2}{2^2}=\dfrac{25x^2-10x+1}{4}\)
\(c,\left(\dfrac{1}{3x-3}\right)\left(\dfrac{1}{3x+3}\right)=\dfrac{1.1}{\left(3x-3\right)\left(3x+3\right)}=\dfrac{1}{9x^2-9}\)
\(d,\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
\(e,\left(\dfrac{1}{4y-2x}\right)^2=\dfrac{1}{\left(4y-2x\right)^2}=\dfrac{1}{16y^2-16xy+4x^2}\)
\(f,\left(2x-y\right)\left(4x^2+2xy+y^2\right)=\left(2x\right)^3-y^3=8x^3-y^3\)
\(g,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
Copy có khác, ko đọc đc j!!! 
ʌl
Câu 3:
1)
a) Ta có: 3x−2=2x−33x−2=2x−3
⇔3x−2−2x+3=0⇔3x−2−2x+3=0
⇔x+1=0⇔x+1=0
hay x=-1
Vậy: x=-1
b) Ta có: 3−4y+24+6y=y+27+3y3−4y+24+6y=y+27+3y
⇔27+2y=27+4y⇔27+2y=27+4y
⇔27+2y−27−4y=0⇔27+2y−27−4y=0
⇔−2y=0⇔−2y=0
hay y=0
Vậy: y=0
c) Ta có: 7−2x=22−3x7−2x=22−3x
⇔7−2x−22+3x=0⇔7−2x−22+3x=0
⇔−15+x=0⇔−15+x=0
hay x=15
Vậy: x=15
d) Ta có: 8x−3=5x+128x−3=5x+12
⇔8x−3−5x−12=0⇔8x−3−5x−12=0
⇔3x−15=0⇔3x−15=0
⇔3(x−5)=0⇔3(x−5)=0
Vì 3≠0
nên x-5=0
hay x=5
Vậy: x=5
a) 3x - 2 = 2x - 3
\(\Leftrightarrow\) 3x - 2 - 2x + 3 = 0
\(\Leftrightarrow\) x + 1 = 0
\(\Rightarrow\) x = -1
b) 3 - 4y + 24 + 6y = y + 27 + 3y
\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0
\(\Leftrightarrow\) -2y = 0
\(\Rightarrow\) y = 0
c)7 - 2x = 22 - 3x
\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0
\(\Leftrightarrow\) -15 + x = 0
\(\Rightarrow\) x = 15
d) 8x - 3 = 5x + 12
\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0
\(\Leftrightarrow\)3x -15 = 0
\(\Leftrightarrow\) 3x = 15
\(\Rightarrow\) x = 5
e) x - 12 + 4x = 25 + 2x - 1
\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0
\(\Leftrightarrow\) 3x - 36 = 0
\(\Leftrightarrow\) 3x = 36
\(\Rightarrow\) x = 12
f ) x + 2x + 3x - 19 = 3x + 5
\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0
\(\Leftrightarrow\)3x - 24 = 0
\(\Leftrightarrow\) 3x = 24
\(\Rightarrow\) x = 8
g) 11+ 8x - 3 = 5x - 3 +x
\(\Leftrightarrow\)8x + 8 = 6x - 3
\(\Leftrightarrow\)8x - 6x = -3 - 8
\(\Leftrightarrow\)2x = -11
\(\Rightarrow\)x = \(-\frac{11}{2}\)
h) 4 - 2x +15 = 9x + 4 -2
\(\Leftrightarrow\)19 - 2x = 7x + 4
\(\Leftrightarrow\)-2x - 7x = 4 - 19
\(\Leftrightarrow\)-9x = -15
\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)
a) \(x^2-y^2-5x-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b) \(5x^3-5x^2y-10x^2+10xy\)
\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x^2-10x\right)\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c) \(x^3-2x^2-x+2\)
\(=\left(x^3-2x^2\right)-\left(x-2\right)\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(-y^2+2xy-x^2+3x-3y\)
\(=-\left(y^2-2xy+x^2\right)+\left(3x-3y\right)\)
\(=-\left(y-x\right)^2+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
g) \(4x^2-8x+3\)
\(=4x^2-6x-2x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-1\right)\)
h) \(2x^2-5x-7\)
\(=2x^2+2x-7x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
k) \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
a) \(2x^3-14x^2-6x\)
b)\(-8x^4y^2+4xy^4-28x^2y^3\)