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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)
\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)
\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)
à xin lỗi mình nhầm dòng cuối
\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)
Để biểu thức trên nhận giá trị dương khi
\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi
a: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)
\(=x^2-2x+1-4x\left(x^2-1\right)+3\)
\(=x^2-2x+4-4x^3+4x\)
\(=-4x^3+x^2+2x+4\)
b: Thay x=-2 vào P, ta được:
\(P=-4\cdot\left(-8\right)+4-4+4=36\)
a: ĐKXĐ: x<>1; x<>-1
b: \(B=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}:\dfrac{2-x^2-x+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{-x^2+1}\)
a, ĐKXĐ:\(\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x\ne-1\\x\ne1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)
b, \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2+x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2-2x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x}{x+1}\)
c, Thay x=2 vào P ta có:
\(P=\dfrac{2x}{x+1}=\dfrac{2.2}{2+1}=\dfrac{4}{3}\)
Bài `1:`
`a)`
Để `P` có nghĩa thì:
`{(x^2-1\ne0),(x+1\ne0),(x-1\ne0):}`
`<=>x\ne+-1`
`b)`
`P=(2x^2)/(x^2-1)+x/(x+1)-x/(x-1)(x\ne+-1)`
`P=(2x^2)/((x-1)(x+1))+(x.(x-1))/((x+1)(x-1))-(x.(x+1))/((x-1)(x+1))`
`P=(2x^2+x^2-x-x^2-x)/((x-1)(x+1))`
`P=(2x^2-2x)/((x-1)(x+1))`
`P=(2x.(x-1))/((x-1)(x+1))=2x/(x+1)`
`c)`
Với `x=2`
`P=(2.2)/(2+1)=4/3`
1) \(A=x^2-6x+9-2x^3+2x=-2x^3+x^2-4x+9\)
2) \(B=x^3-3x+2x^2-6-x^3+1=2x^2-3x-5\)
Ta có: \(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{x^2-1}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4x+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4}{x+1}\)
\(\dfrac{x-1}{x+1}\) - \(\dfrac{x+1}{-\left(1-x\right)}\) - \(\dfrac{4}{\left(1-x\right)\left(1+x\right)}\) MTC: -(1 - x)(1 + x)
= \(\dfrac{-\left(x-1\right)^2}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{\left(x+1\right)^2}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{-4}{\text{-(1 - x)(1 + x)}}\)
= \(\dfrac{-x^2+2x-1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{x^2+2x+1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{-4}{\text{-(1 - x)(1 + x)}}\)
= \(\dfrac{-2x^2+2}{\text{-(1 - x)(1 + x)}}\) = \(\dfrac{-2\left(x^2-1\right)}{\text{-(1 - x)(1 + x)}}\) = \(\dfrac{2\text{(x - 1)(1 + x)}}{\text{(1 - x)(1 + x)}}\) = \(\dfrac{2x-2}{1-x}\)
\(B=\frac{x}{x-1}+\frac{1}{x+1}+\frac{2}{x^2-1}\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+x+x-1+2}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)
bạn có thể tìm các giá trị x để giá trị của biểu thức b là cấc số nguyên tố nhỏ hơn 10 dc ko