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a/ (x2-1)(x2+2x)
=x4+2x3-x2-2x
b/ (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+x-6
c/ (x+3)(x2+3x-5)
=x3+3x2-5x+3x2+9x-15
=x3+6x2+4x-15
d/ (x+1)(x2-x+1)
=x3+1 dùng HĐT
e/ (2x3-3x-1)(5x+2)
=10x4-15x2-5x+4x3-6x-2
=10x4+4x3-15x2-11x-2
f/ (x2-2x+3)(x-4)
=x3-2x2+3x-4x2+8x-12
=x3-6x2+11x-12
a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+3x-x^2-6+2x\)
\(=-6x^3+17x^2+5x-6\)
c) Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+1\)
e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)
\(=10x^4+4x^3-15x^2-6x-5x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)
\(=x^3-4x^2-2x^2+8x+3x-12\)
\(=x^3-6x^2+11x-12\)
g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)
\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)
\(=7x^2+16x-1-x^2+7x-12\)
\(=6x^2+23x-23\)
h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)
\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)
\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)
\(=-5x^3+26x^2-28x-2\)
1.a) \(\Leftrightarrow\) 3x+10-2x =0
\(\Leftrightarrow\text{ 3x-2x=-10}\)
\(\Leftrightarrow x=-10\)
b) coi lại có thiếu ngoặc ko nhé
cứ nhân vào dấu ngoặc rồi làm như thường
. Làm như bt thôi mà bạn :) , áp dụng hằng đẳng thức cùng vs nhân chi tr - cộng trừ sau là ra mà bạn !!!
a, (x4-2x3+2x-1):(x2-1) = \(\frac{\left(x^4-1\right)-\left(2x^3-2x\right)}{x^2-1}\)
= \(\frac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\) =\(\frac{\left(x^2-1\right)\left(x^2+1-2x\right)}{x^2-1}\)
= \(x^2+1-2x\)= \(\left(x-1\right)^2\)
b, (8x3-6x2-5x+3):((4x+3)
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) 
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, \(\left(x+1\right)\left(x-2\right)=x^2-2x+x-2=x^2-x-2\)
b, \(-7x^2\left(3x-4y\right)=-21x^3+28x^2y\)
c, \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)
\(=x^2+2x-8-x^2+6x-9=8x-17\)
Mấy bài dài dài kia tí mình làm cho :)
( x - 1 )3 - x( x - 2 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 4x + 4 ) + 1
= x3 - 3x2 + 3x - x3 + 4x2 - 4x
= x2 - x = x( x - 1 )
2x( 3x + 2 ) - 3x( 2x + 3 )
= 6x2 + 4x - 6x2 - 9x
= -5x
( x + 2 )3 + ( x - 3 )2 - x2( x + 5 )
= x3 + 6x2 + 12x + 8 + x2 - 6x + 9 - x3 - 5x2
= 2x2 + 6x + 17
( 2x + 3 )( x - 5 ) + 2x( 3 - x ) + x - 10
= 2x2 - 7x - 15 + 6x - 2x2 + x - 10
= -25
( x + 5 )( x2 - 5x + 25 ) - x( x - 4 )2 + 16x
= x3 + 53 - x( x2 - 8x + 16 ) + 16x
= x3 + 125 - x3 + 8x2 - 16x + 16
= 8x2 + 125
( -x - 2 )3 + ( 2x - 4 )( x2 + 2x + 4 ) - x2( x - 6 )
= -x3 - 6x2 - 12x - 8 + 2x3 - 16 - x3 + 6x2
= -12x - 24 = -12( x + 2 )
Tương tự ...
a, \(\left(x-1\right)^3-x\left(x-2\right)^2+1=x^3-3x^2+3x-1-x^3+4x^2-4x+1=x^2-x\)
b, \(2x\left(3x+2\right)-3x\left(2x+3\right)=6x^2+4x-6x^2-9x=-5x\)
c, \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)=x^3+6x^2+12x+8+x^2+6x+9-x^3-5x^2=2x^2+18x+17\)