rút gọn \(\dfrac{3x+3}{3x}\)
Rút gọn A = \(\dfrac{3x}{x-1}\)+\(\dfrac{2}{x+1}\)+\(\dfrac{3-3x-2x^2}{x^2-1}\)
\(A=\dfrac{3x}{x-1}+\dfrac{2}{x+1}+\dfrac{3-3x-2x^2}{x^2-1}.\) \(\left(ĐKXĐ:x\ne1;x\ne-1\right).\)
\(A=\dfrac{3x\left(x+1\right)+2\left(x-1\right)+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.\)
\(A=\dfrac{3x^2+3x+2x-2+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.\)
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}.\)
1) rút gọn
a) \(\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\)
b) \(\dfrac{x^3+3x^2-2}{x^3+3x+4}=\)
\(b,\dfrac{x^3+3x^2-2}{x^3+3x+4}=\dfrac{x^3+x^2+2x^2+2x-2x-2}{x^3+x^2-x^2-x+4x+4}\\ =\dfrac{x^2\left(x+1\right)+2x\left(x+1\right)-2\left(x+1\right)}{x^2\left(x+1\right)-x\left(x+1\right)+4\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^2+2x-2\right)}{\left(x+1\right)\left(x^2-x+4\right)}=\dfrac{x^2+2x-2}{x^2-x+4}\)
\(a,\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\dfrac{\left(x^2-y^2\right)+\left(3x-3y\right)}{x^2-y^2}\\ =\dfrac{\left(x-y\right)\left(x+y\right)+3\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y+3\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+3}{x+y}\)
rút gọn biểu thức:
a) \(\dfrac{2x^{2^{ }}-2x}{x-1}\)
b)\(\dfrac{x^{2^{ }}+2x+1}{3x^2+3x}\)
c)\(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)
a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)
\(=\dfrac{2x\left(x-1\right)}{x-1}\)
=2x
b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)
\(=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}\)
\(=\dfrac{x+1}{3x}\)
c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)
\(=\dfrac{x}{3\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+3}{3\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+4}{3x^2-3}\)
a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )
b,\(\dfrac{x^2+2x+1}{3x^2+3x}=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{x+1}{3x}\) ( đk : \(x\ne-1\) )
c
=
Rút gọn: M = \(\dfrac{1}{3x-2}-\dfrac{4}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
M= 1/ 3x-2 - 4/ 3x +2 - 3x-6/4-9x^2
= 3x+2 - 12x + 8 + 3x-6
= -6x +4
`M=1/(3x-2)-4/(3x+2)-(3x-6)/(4-9x^2)(x ne +-2/3)`
`=(3x+2-4(3x-2)+3x+6)/(9x^2-4)`
`=(-6x+16)/(9x^2-4)`
1) Rút gọn
a) (3x - 2)2 - (1+ 5x)2
b) (3x + 4)(3x - 4) - (5 - x)2
c) (\(\dfrac{1}{2}\)x + 4)2 - (\(\dfrac{1}{2}\)x + 3)(\(\dfrac{1}{2}\)x - 3)
a) (3x - 2)2 - (1 + 5x)2
= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)
= (-2x - 3)(8x - 1)
b) (3x + 4)(3x - 4) - (5 - x)2
= (3x)2 - 42 - (25 - 10x + x2)
= 9x2 - 16 - 25 + 10x - x2
= 8x2 + 10x - 41
c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)
\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)
\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)
\(=4x+25\)
a: =9x^2-12x+4-25x^2-10x-1
=-16x^2-22x+3
b: =9x^2-16-x^2+10x-25
=8x^2+10x-41
c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)
\(\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{3x^2}\)
giúp mik rút gọn câu này vs
ĐK: x\(\ne\){-3;0;3}.
\(\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{3x^2}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{3x^2}=\dfrac{1}{3x}\).
Rút gọn M và A sau đây :
M= \(\left(\dfrac{x}{x+3}+\dfrac{3-x}{x+3}.\dfrac{x^2+3x+9}{x^2-9}\right)\)
A= \(\left(\dfrac{3x}{1-3x}-\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
1 Rút gọn
\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}\)
\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)+\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy+1}\)
\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-a\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\)
\(\sqrt{3x+2}-3x\) vs x ≥ \(\dfrac{-2}{3}\)
rút gọn nhoa !!!
mk cảm ơn
Biểu thức này không rút gọn được nữa.
Vì giờ bạn muốn rút gọn thì bạn phải rút gọn được $\sqrt{3x+2}$. Mà $\sqrt{3x+2}$ có cái gì để rút gọn nữa đâu??