Thu gọn biểu thức A = 3x + 21 x 2 − 9 + 2 x + 3 − 3 x − 3 ta được?
A. − 2 x − 3
B. 2x (x − 3)(x + 3)
C. 2 x + 3
D. 2 x − 3
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CM
Cao Minh Tâm
16 tháng 7 2017
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CM
12 tháng 7 2017
x + 2 x - 3 = x - x + 3 x - 3 = x ( x - 1) + 3( x - 1) = ( x - 1)( x + 3)
a) Với điểu kiện x ≥ 0; x ≠ 1 ta có:
MT
0
NT
0
MK
18 tháng 12 2017
A\(\dfrac{x^3-9x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)=\dfrac{x\left(x^2-9\right)}{2x+3}\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x+3\right)\left(x-3\right)}\right)=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}\left(\dfrac{\left(x+3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{x.x}{\left(x+3\right)\left(x-3\right)x}\right)=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}\left(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}-\dfrac{x^2}{\left(x+3\right)\left(x-3\right)x}\right)\)
\(=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(x+3+x\right)\left(x+3-x\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(2x+3\right).3}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)\left(2x+3\right).3}{\left(2x+3\right)x\left(x-3\right)\left(x+3\right)}=3\)
NH
1
10 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`(3x^2 + 2x)*(x^2 - 3) + (4 - x^3) * (3x+2)`
`= 3x^2(x^2 - 3) + 2x(x^2 - 3) + 4(3x+2) - x^3(3x+2)`
`= 3x^4 - 9x^2 + 2x^3 - 6x + 12x + 8 - 3x^4 - 2x^3`
`= (3x^4 - 3x^4) + (2x^3 - 2x^3) - 9x^2 + (-6x+12x) + 8`
`= -9x^2 + 6x + 8`
2 tháng 10 2018
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
D
0
NQ
1
Y
25 tháng 6 2023
\(A=\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}+\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\left(dk:x\ne\pm3,x\ne2\right)\)
\(=\dfrac{3x-x^2-9+x^2}{9-x^2}:\left(\dfrac{9-x^2}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{x-2}-\dfrac{x+2}{x+3}\right)\)
\(=\dfrac{3x-9}{9-x^2}:\dfrac{9-x^2-\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-\left(x^2-9\right)-\left(x^2-4\right)}\)
\(=-\dfrac{3}{x+3}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-x^2+9-x^2+4}\)
\(=\dfrac{-3\left(x-2\right)}{22-3x^2}\)
\(=\dfrac{-3x+6}{22-3x^2}\)
Vậy \(A=\dfrac{-3x+6}{22-3x^2}\) với \(x\ne\pm3,x\ne2\)