Làm tính chia phân thức:
a) 1 − 9 x 2 x 2 + 4 x : 2 − 6 x 3 x với x ≠ − 4 ; x ≠ 0 và x ≠ 1 3 ;
b) y 3 + 8 y − 1 : ( y 2 − 2 y + 4 ) với y ≠ 1 .
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`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
A= x(x - 2 )( x + 2 ) - ( x- 3 )( x^2 + 3x + 9 )
= x ( x^2 - 4 ) - (x^3 - 27 )
= x^3 - 4x - x^3 + 27
= 27 - 4x
Thay x= 1/4 vào A ta có :
A = 27 - 4.1/4 = 27 - 1 = 6
\(\dfrac{3-3x}{x^2-9}\cdot\dfrac{x-3}{x-1}\\ =\dfrac{3\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =\dfrac{-3\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =-\dfrac{3}{x+3}\\ \dfrac{6x+4}{x^2-4}\cdot\dfrac{x^2-2x}{3x+2}\\ =\dfrac{2\left(3x+2\right)x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}\\ =\dfrac{2x}{x+2}\)
a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A
\(A=x^6-\left(x+1\right)^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(A=1\)
b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B
\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)
\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)
\(B=1\)
Chúc bạn học tốt!!!
`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`
`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`
`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`
`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`
`=1/(sqrtx-2)`
\(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)
\(=\left(\dfrac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\sqrt{x}+3\right)=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`
`= 2xy`.
Thay `x = 2/3; y = -3/4` vào BT:
`2 . 2/3 . -3/4 = -1.`
`b, x(x-2y) - y(y^2-2x)`
`= x^2 - 2xy - y^3 + 2xy`
`= x^2 - y^3`
Thay `x = 5; y =3` vào BT:
`= 5^2 - 3^3 = 25 - 27 = -2`
a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)
\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)
\(=2xy\)
Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:
\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)
b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)
\(=x^2-2xy-y^3+2xy\)
\(=x^2-y^3\)
Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)
a) Áp dụng quy tắc chia hai phân thức và rút gọn ta có kết quả là 3 ( 1 + 3 x ) 1 ( x + 4 )
b) Tương tự a), ta tính được kết quả là y + 2 y − 1