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31 tháng 10 2019
a) \(P=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
\(P=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x+1\right)}\)
31 tháng 10 2019
b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x^2+4x-5}{2\left(x+5\right)}\)
YN
22 tháng 2 2022
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
20 tháng 10 2016
A= (2x-1)2-(2x+3)(x-2)-2(x+2)(x+5)
A= 4x2-4x+1-(2x2-x-6)-2(x2+7x+10)
A=4x2-4x+1-2x2+x+6-2x2-14x-20
A= -17x-13
Thay x= -3, ta có:
A= -17.3-13=-51-13=-64
6 tháng 1 2018
\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)
\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)
\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)
\(P=\frac{x^4+1}{2x+1}\)
Vậy \(P=\frac{x^4+1}{2x+1}\)
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow A=\left(\frac{-1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\left[\frac{-x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-2}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}\)
\(\Leftrightarrow A=\frac{2\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=1\)
vậy \(A=1\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
A
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}\cdot\frac{^2-1}{1-2x}=\frac{2}{1-2x}\)ĐK: x khác 1/2
31 tháng 10 2020
Bài làm
Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha
\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)
\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)
\(A=\frac{\left(x+2\right)2\left(x^2-4\right)}{2\left(2x^2+2\right)}\)
\(A=\frac{2\left(x+2\right)\left(x-2\right)\left(x+2\right)}{4\left(x^2+1\right)}\)
\(A=\frac{\left(x+2\right)^2\left(x-2\right)}{2\left(x^2+1\right)}\)
:>>> Chả biết đúng không nữa nhưng số to quá :>>
12 tháng 11 2021
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
\(=\left(x+5+2x+5\right)^2=\left(3x+10\right)^2\)