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QA
5 tháng 7 2021
Trả lời:
a, ( x + 1 )2 + ( x - 2 ) ( x + 3 ) - 4x
= x2 + 2x + 1 + x2 + 3x - 2x - 6 - 4x
= 2x2 - x - 5
b, ( x - 2 )2 + ( x + 1 )2 + 2 ( x - 2 ) ( - 1 - x )
= x2 - 4x - 4 + x2 + 2x + 1 + ( 2x - 4 ) ( - 1 - x )
= 2x2 - 2x - 3 - 2x - 2x2 + 4x + 4x
= 4x - 3
TA
5 tháng 7 2021
a) \(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x\)
\(=\left(x^2+2x+1\right)+\left(x^2+x-6\right)-4x\)
\(=x^2+2x+1+x^2+x-6-4x\)
\(=2x^2-x-5\)
b) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x^2-4x+4\right)+\left(x^2+2x+1\right)+\left(2x-4\right)\left(-1-x\right)\)
\(=x^2-4x+4+x^2+2x+1+\left(-2x-2x^2+4+4x\right)\)
\(=x^2-4x+4+x^2+2x+1-2x-2x^2+4+4x\)
\(=9\)
KC
30 tháng 12 2017
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)
TK
0
A
7 tháng 8 2019
a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
A
7 tháng 8 2019
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)
16 tháng 9 2020
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2. (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
B
16 tháng 9 2020
Bài làm :
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2) (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
PT
1
CM
23 tháng 3 2017
(x - 1/2 )(x + 1/2 )(4x - 1)
= ( x 2 + 1/2 x - 1/2 x - 1/4 )(4x - 1)
= ( x 2 - 1/4 )(4x - 1)
= 4 x 3 – x 2 – x + 1/4
6 tháng 8 2020
a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7
\(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)\)
\(=x^2+\frac{1}{2}x-\frac{1}{2}x-\frac{1}{4}\left(4x-1\right)\)
\(=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)\)
\(=4x^3-x^2-x+\frac{1}{4}\)
ta có: (x - 1/2)(x + 1/2)(4x - 1) = 0
=> (x2 - 1/4)(4x - 1) = 0
=> \(\hept{\begin{cases}x^2-\frac{1}{4}=0\\4x-1=0\end{cases}}\)
=> \(\hept{\begin{cases}x^2=\frac{1}{4}\\4x=1\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{1}{2}hoặc-\frac{1}{2}\\x=\frac{1}{4}\end{cases}}\)
ok nhé!! 364565467567776892512352534534534564654645645645756756