Cho P = (\(\dfrac{2x}{2x^2-5x+3}\)-\(\dfrac{5}{2x-3}\))\(\div\left(3-\dfrac{2}{1-x}\right)\)
a, Rút gọn P
b,Tính P với \(\left|3x-2\right|+1=5\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
9 tháng 5 2023
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
25 tháng 2 2021
`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`
`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`
`=(y(y-2x))/3`
`b,(x^2-y^2)/(x^2-y^2+xz-yz)`
`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`
`=(x+y)/(x+y+z)`
`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`
`=(-(x^2-3x+x-3))/((x-1)(x+1))`
`=(-x(x-3)+x-3)/((x-1)(x+1))`
`=((x-3)(1-x))/((x-1)(x+1))`
`=(3-x)/(1+x)`
`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`
`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`
`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`
`=(3x^2-4x+1)/(2x^2+x-3)`
`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`
`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`
`=(3x-1)/(2x+3)`
25 tháng 2 2021
a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)
\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)
\(=\dfrac{y\left(y-2x\right)}{3}\)
23 tháng 12 2021
\(A=\left(\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{x^2+x-2+x^2-x+2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x^2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x\left(x^2-2\right)\left(x-3\right)}{2x\left(x-2\right)\cdot\left(x+2\right)^2}=\dfrac{\left(x^2-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)^2}\)
19 tháng 6 2021
a) đk: x khác 1; \(\dfrac{3}{2}\)
\(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)
= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)
= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)
b) Có \(\left|3x-2\right|+1=5\)
<=> \(\left|3x-2\right|=4\)
<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)
TH1: Thay x = 2 vào P, ta có:
P = \(\dfrac{-1}{2.2-3}=-1\)
TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:
P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)
c) Để P > 0
<=> \(\dfrac{-1}{2x-3}>0\)
<=> 2x - 3 <0
<=> x < \(\dfrac{3}{2}\) ( x khác 1)
d) P = \(\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)
<=> 2x - 3 = x2 - 6
<=> x2 - 2x - 3 = 0
<=> (x-3)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)
17 tháng 7 2021
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-1\right):\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}+1\right)+\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}-1\right)-2x+1}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}:\left(\dfrac{2x-1+\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)-\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}\right)\)
\(=\dfrac{x\sqrt{2}+\sqrt{x}+\sqrt{2x}+1+2x-\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{2x-1}:\dfrac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{x}+2}{-2-2\sqrt{x}}\)
12 tháng 10 2021
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
a, \(P=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{1-x}\right)\)ĐK : \(x\ne1;\dfrac{3}{2};\dfrac{1}{3}\)
\(=\left(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(3+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\right):\left(\dfrac{3x-3+2}{x-1}\right)\)
\(=\dfrac{\left(5-3x\right)\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)\left(3x-1\right)}=\dfrac{5-3x}{\left(2x-3\right)\left(3x-1\right)}\)
b, \(\left|3x-2\right|+1=5\Leftrightarrow\left|3x-2\right|=4\)
TH1 : \(3x-2=4\Leftrightarrow x=2\)
TH2 : \(3x-2=-4\Leftrightarrow x=-\dfrac{2}{3}\)
Với \(x=2\Rightarrow P=\dfrac{5-6}{5}=-\dfrac{1}{5}\)
Với \(x=-\dfrac{2}{3}\Rightarrow P=\dfrac{5+2}{\left(-\dfrac{4}{3}-3\right)\left(-3\right)}=\dfrac{7}{-\dfrac{13}{3}.\left(-3\right)}=\dfrac{7}{13}\)
a) Ta có: \(P=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{1-x}\right)\)
\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3\left(1-x\right)-2}{1-x}\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x-2}{1-x}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{1-x}{-3x+1}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-1}\)
\(=\dfrac{-3x+5}{2x-3}\)