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`

\(a,P=2x\left(-3x+2\right)-\left(x+2\right)^2+8x^2-1\)

\(=-6x^2+4x-\left(x^2+4x+4\right)+8x^2-1\)

\(=-6x^2+4x-x^2-4x-4+8x^2-1\)

\(=\left(-6x^2-x^2+8x^2\right) +\left(4x-4x\right)+\left(-4-1\right)\)

\(=x^2-5\)

Vậy \(P=x^2-5\).

\(b,\) Ta có: \(P=x^2-5\)

Thay \(x=3\) vào \(P\), ta được:

\(P=3^2-5=9-5=4\)

Vậy \(P=4\) khi \(x=3\).

\(c,\) Có: \(P=-1\)

\(\Leftrightarrow x^2-5=-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(P=-1\) khi \(x\in\left\{2;-2\right\}\).

#\(Toru\)

a) P = 2x(-3x + 2) - (x + 2)² + 8x² - 1

= -6x² + 4x - x² - 4x - 4 + 8x² - 1

= (-6x² - x² + 8x²) + (4x - 4x) + (-4 - 1)

= x² - 5

b) Thay x = 3 vào P, ta được:

P = 3² - 5

= 4

c) Để P = -1 thì x² - 5 = -1

x² = -1 + 5

x² = 4

x = 2 hoặc x = -2

Vậy x = 2; x = -2 thì P = -1

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)

a: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

mà n là số nguyên

nên \(n\in\left\{0;-1;1\right\}\)

b: \(\Leftrightarrow10n^2-10n+11n-11+1⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1\right\}\)

hay \(n\in\left\{2;0\right\}\)

c: \(\Leftrightarrow x^4-x^3+5x^2+x^2-x+5+n-5⋮x^2-x+5\)

=>n-5=0

hay n=5

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