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a. Để P được xđ thì MT phải khác 0.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9\ne0\\x^2+3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ne0\\x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.\)
b. \(P=\left(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{1}{x}\)
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
\(a,P=2x\left(-3x+2\right)-\left(x+2\right)^2+8x^2-1\)
\(=-6x^2+4x-\left(x^2+4x+4\right)+8x^2-1\)
\(=-6x^2+4x-x^2-4x-4+8x^2-1\)
\(=\left(-6x^2-x^2+8x^2\right) +\left(4x-4x\right)+\left(-4-1\right)\)
\(=x^2-5\)
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: \(P=\dfrac{2}{3x+2}-\dfrac{1}{3x-2}+\dfrac{4}{9x^2-4}\)
\(=\dfrac{6x-4-3x-2+4}{\left(3x+2\right)\left(3x-2\right)}=\dfrac{3x-2}{\left(3x+2\right)\left(3x-2\right)}=\dfrac{1}{3x+2}\)
a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)
b: Thay x=11 vào M, ta được:
\(M=\dfrac{8}{11-3}=1\)
a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)
\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)
b) Thay \(x=11\left(TM\right)\) vào biểu thức M:
\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)
a, P xác định khi \(x^3-8\ne0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\ne0\)
\(\Leftrightarrow x\ne2\left(\text{Vì }x^2+2x+4>0\right)\)
b, \(P=\dfrac{3x^2+6x+12}{x^3-8}=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c, \(x=\dfrac{4001}{2000}\Rightarrow P=\dfrac{3}{\dfrac{4001}{2000}-2}=6000\)
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
a) ĐKXĐ `x + 3 ne 0 ` và `x -3 ne 0` và ` 9 -x^2 ne 0`
`<=> x ne -3 ` và `x ne 3` và `(3-x)(3+x) ne 0`
`<=> x ne -3` và `x ne 3`
b) Với `x ne +-3` ta có:
`P= 3/(x+3) + 1/(x-3)- 18/(9-x^2)`
`P= [3(x-3)]/[(x-3)(x+3)] + (x+3)/[(x-3)(x+3)] + 18/[(x-3)(x+3)]`
`P= (3x-9)/[(x-3)(x+3)] + (x+3)/[(x-3)(x+3)] + 18/[(x-3)(x+3)]`
`P= (3x-9+x+3+18)/[(x-3)(x+3)]`
`P= (4x +12)/[(x-3)(x+3)]`
`P= (4(x+3))/[(x-3)(x+3)]`
`P= 4/(x-3)`
Vậy `P= 4/(x-3)` khi `x ne +-3`
c) Để `P=4`
`=> 4/(x-3) =4`
`=> 4(x-3) = 4`
`<=> 4x - 12=4`
`<=> 4x = 16
`<=> x= 4` (thỏa mãn ĐKXĐ)
Vậy `x=4` thì `P =4`
a) P xác định <=> \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
<=>\(x\ne\pm3\)
b)Với \(x\ne\pm3\)
\(P=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)
\(=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x-3\right)+\left(x+3\right)+18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4}{x-3}\)
c)Với \(x\ne\pm3\)
P=4 <=>\(\dfrac{4}{x-3}=4\)
<=>\(4x-12=4\)
<=>\(4x=16\)
<=>x=4(tm)
Vậy x=4
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
