Rút gọn phân thức sau: x 3 - 27 9 - 6 x + x 2 .
A. - ( x 2 + 3 x + 9 ) 3 - x
B. x 2 + 3 x + 9 3 - x
C. x 2 + 3 x + 9 3 + x
D. x 2 + 3 x 3 - x
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a) Ta có: \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)\)
\(=x^3-6x^2+12x-8+\left(x-3\right)\left(x^2+3\right)\)
\(=x^3-6x^2+12x-8+x^3+3x-3x^2-9\)
\(=2x^3-9x^2+15x-17\)
b) Ta có: \(x\left(x-14\right)-10\left(x-1\right)^2\)
\(=x^2-14x-10\left(x^2-2x+1\right)\)
\(=x^2-14x-10x^2+20x-10\)
\(=-9x^2+6x-10\)
c) Ta có: \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=2x^2+4x-\left(x^2-4\right)\)
\(=2x^2+4x-x^2+4\)
\(=x^2+4x+4\)
d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3-27\right)\)
\(=x^3-27-x^3+27\)
=0
\(A=\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}+\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\left(dk:x\ne\pm3,x\ne2\right)\)
\(=\dfrac{3x-x^2-9+x^2}{9-x^2}:\left(\dfrac{9-x^2}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{x-2}-\dfrac{x+2}{x+3}\right)\)
\(=\dfrac{3x-9}{9-x^2}:\dfrac{9-x^2-\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-\left(x^2-9\right)-\left(x^2-4\right)}\)
\(=-\dfrac{3}{x+3}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-x^2+9-x^2+4}\)
\(=\dfrac{-3\left(x-2\right)}{22-3x^2}\)
\(=\dfrac{-3x+6}{22-3x^2}\)
Vậy \(A=\dfrac{-3x+6}{22-3x^2}\) với \(x\ne\pm3,x\ne2\)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`
`=x^3-3^3-(9x^3+27x^2-9x-27)`
`=x^3-3^3-9x^3-27x^2+9x+27`
`=-8x^3-27x^2+9x`
b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`
`=x^3-2^3-x(x^2-9)`
`=x^3-8-x^3+9x`
`=9x-8`
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)
\(=x^3-27-\left(9x^3+27x^2-9x-27\right)\)
\(=x^3-27-9x^3-27x^2+9x+27\)
\(=-8x^3-27x^2+9x\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=x^3-8-x\left(x^2-9\right)\)
\(=x^3-8-x^3+9x\)
\(=9x-8\)
\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)
\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)
\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)
\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)
a) (x-3)(x2+3x+9)-(x2-1)(x+27)
=(x3-27)-(x3+27x2-x-27)
=x3-27-x3-27x2+x+27
=-27x2+x
=x(-27x+1)
b) (3-x)3-(x+3)(x2-3x+9)
=27-27x+9x2-x3-x3-27
=-2x3+9x2-27x
=x(-2x+9x-27)
c) (x-2)(x2+2x+4)-x(x-3)(x+3)
=x3-8-x(x2-9)
=x3-8-x3+9x
=9x-8
#H
a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)
Tử số= 2^19.9^3.3^3 + 5.3.2^9.2^9.9^4
= 2^19.9^3.3^3 + 5.3.2^18.9.9^3
= 2^19.9^3.3.3^2 + 5.3.2^18.3^2.9^3
= 2^18.9^3.3^2(2 + 5.) (đặt nhân tử chung)
=7.2^18.9^3.3^2
=7.2^18.9.9.9.3^2
=7.2^18.3^2.3^2.3^2.3^2
=7.2^18.3^8
Mẫu số= 6^9.2^10 + 6^10.2^10
= 6^9.2^10 + 6^10.2^10
=6^9.2^10(1+6)
=7.6^9.2^10.
=7.2^9.3^9.2^10
=7.2^19.3^9
Lấy tử số chia mẫu số ta được : 1/2.3 = 1/6
\(-\left(x+3\right)^3+\left(x+9\right)\left(x^2+27\right)=-\left(x^3+27+9x^2+27x\right)+x^3+27x+9x^2+243\)
\(=-x^3-27-9x^2-27x+x^3+27x+9x^2+243=243-27=216\)
nha . Cảm ơn . Chúc bạn học tốt
a) ĐKXĐ:
\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)
b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)
\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)
c) Thay x = - 1 vào A ta có:
\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)
Chọn đáp án A