Biết \(\dfrac{x}{x^2-9}+\dfrac{2}{x^2+6x+9}=0vớix\ne3.Tìmx\)
Những câu hỏi liên quan
1.rút gọn biểu thuc P=\(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}\) với x\(\ne-3vàx\ne3\)
2.thực hiện phép tính \(\left(2x^4-3x^3-3x^2+6x-1\right):\left(x^2-2\right)\)
\(\left(15x^4y^6-12^3y^4-18x^2y^3\right):\left(-6x^2y^2\right)\)
1) \(\dfrac{1}{x^2+6x+9}+\dfrac{1}{6x-x^2+9}+\dfrac{x}{x^2-9}\) 2) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\) 3) \(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\)
BÀI 1 : RÚT GỌN CÁC BIỂU THỨC SAU .
a, \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
b, \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
c, \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d, \(\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Rút gọn M và A sau đây :
M= \(\left(\dfrac{x}{x+3}+\dfrac{3-x}{x+3}.\dfrac{x^2+3x+9}{x^2-9}\right)\)
A= \(\left(\dfrac{3x}{1-3x}-\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
câu 53:điều kiện để phân thức \(\dfrac{7x^2-x}{x^2-9}\) được xác định khi :
a.\(x\ne3\) b,\(x\ne9\) c,\(x\ne3;x\ne-3\) d,\(x\ne-9vàx\ne9\)
GIẢI THÍCH VÌ SAO LẠI RA KẾT QUẢ ĐÓ
Phân thức `(7x^2 - x)/(x^2 - 9)` được xác đinh khi
`x^2-9≠0`
`(x-3).(x+3)≠0`
`=> x≠3; x ≠ -3`
`=> C`
Đúng 3
Bình luận (0)
thực hiện phép tính
\(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(\dfrac{6}{x-3}-\dfrac{2x-16}{x^2-9}-\dfrac{4}{x+3}\)
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
Đúng 0
Bình luận (0)
18 tháng 7 2021 lúc 9:34
\(\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}\) =\(\dfrac{a}{b-x}\)
Tìm a,ba,b để được đẳng thức đúng
Hãy chứng tỏ các phân thức sau bằng nhaua/ dfrac{x+3}{2x-5}dfrac{x^2+3x}{2x^2-5x}b/ dfrac{3-x}{x+3}dfrac{x^2-6x+9}{9-x^{ }}c/ dfrac{x^3+64}{left(3-xright)left(x^2-4x+16right)}dfrac{x-4}{x-3}d/ dfrac{x^3+6x^2-x-30}{x^3+3x^2-25x-75}dfrac{x-2}{x-5}AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE Ạ
Đọc tiếp
Hãy chứng tỏ các phân thức sau bằng nhau
a/ \(\dfrac{x+3}{2x-5}=\dfrac{x^2+3x}{2x^2-5x}\)
b/ \(\dfrac{3-x}{x+3}=\dfrac{x^2-6x+9}{9-x^{ }}\)
c/ \(\dfrac{x^3+64}{\left(3-x\right)\left(x^2-4x+16\right)}\)\(=\dfrac{x-4}{x-3}\)
d/ \(\dfrac{x^3+6x^2-x-30}{x^3+3x^2-25x-75}=\dfrac{x-2}{x-5}\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE Ạ
\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)
Đúng 1
Bình luận (0)
Bài 1: Rút gon
a) B=\(\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right).\dfrac{3x^2-9x}{x^2+6x+9}\)
b) A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)
\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)
b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{-6}{x-2}\)
Đúng 2
Bình luận (0)