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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. \(2a^2+5ab-3b^2-7b-2\)
\(=\left(2a^2+6ab+2a\right)-\left(ab+3b^2+b\right)-\left(2a+6b+2\right)\)
\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)
\(=\left(2a-b-2\right)\left(a+3b+1\right)\)
b. \(2x^2-7xy+x+3y^2-3y\)
\(=\left(2x^2-xy\right)-\left(6xy-3y^2\right)+\left(x-3y\right)\)
\(=x\left(2x-y\right)-3y\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
c. \(6x^2-xy-2y^2+3x-2y\)
\(=\left(6x^2+3xy\right)-\left(4xy-2y^2\right)+\left(3x-2y\right)\)
\(=3x\left(2x+y\right)-2y\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y+1\right)\)
giải
A=(3x-5)(2x+11)-(2x+3)(3x+7)
=6x^2+33x-10x-55-(6x^2+14x+9x+21)
=6x^2+33x-10x-55-6x^2-14x-9x-21
= -76
vậy biểu thức không phụ thuộc vào biến x,y
B=(2x+3)(4x^2-6x+9)-2(4x^3-1)
=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2
=29
vậy biểu thức không phụ thuộc vào biến x
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 = \(\frac{15}{9}\) = \(\frac{5}{3}\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
Bài 7: Phân tích đa thức thành nhân tử
a) Ta có: \(a^2-b^2-2a+2b\)
\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2\right)\)
b) Ta có: \(3x-3y-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
c) Ta có: \(16-x^2+4xy-4y^2\)
\(=16-\left(x^2-4xy+4y^2\right)\)
\(=16-\left(x-2y\right)^2\)
\(=\left(4-x+2y\right)\left(4+x-2y\right)\)
d) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(5-x-4y\right)\left(3x+2y+3\right)\)
e) Ta có: \(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
f) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)
\(=\left(x+3+x-3\right)\left[\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\right]\)
\(=2x\cdot\left[x^2+6x+9-\left(x^2-9\right)+x^2-6x+9\right]\)
\(=2x\cdot\left(2x^2+18-x^2+9\right)\)
\(=2x\cdot\left(x^2+27\right)\)
g) Ta có: \(9x^2-3xy+y-6x+1\)
\(=\left(9x^2-6x+1\right)-y\left(3x-1\right)\)
\(=\left(3x-1\right)^2-y\left(3x-1\right)\)
\(=\left(3x-1\right)\left(3x-1-y\right)\)
h) Ta có: \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)