Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{4.\left(x+3\right)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4.\left(x+3\right)}{\left(3x-1\right)}\cdot\frac{\left(3x-1\right)}{x^2+3x}=\frac{4.\left(x+3\right)}{x.\left(x+3\right)}=\frac{4}{x}\)
\(a,\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)
\(=\frac{-x-2}{1-x}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)
\(=\frac{-x-2}{1-x}+\frac{-\left(x-9\right)}{1-x}+\frac{-\left(x-9\right)}{1-x}\)
\(=\frac{-x-2-x+9-x+9}{1-x}=\frac{-3x+16}{1-x}\)
Câu b,c mk chưa học, bn thông cảm
Còn câu a, nếu sai thì xin lượng thứ :))
Lời giải:
a)
\(\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}=\frac{x+2}{x-1}-\frac{2(x-9)}{1-x}\)
\(=\frac{x+2}{x-1}+\frac{2(x-9)}{x-1}=\frac{x+2+2(x-9)}{x-1}=\frac{3x-16}{x-1}\)
b)
\(\frac{x^2-9y^2}{x^2y}: \frac{xz-3yz}{3xy}=\frac{x^2-9y^2}{x^2y}.\frac{3xy}{xz-3yz}\)
\(=\frac{(x-3y)(x+3y)}{x^2y}.\frac{3xy}{z(x-3y)}=\frac{3(x+3y)}{xz}\)
c) \(\frac{4(x+3)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4(x+3)}{3x-1}.\frac{3x-1}{x^2+3x}=\frac{4(x+3)}{x^2+3x}=\frac{4(x+3)}{x(x+3)}=\frac{4}{x}\)
b: \(=\dfrac{-1}{x\left(5x-1\right)}-\dfrac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)
\(=\dfrac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}\)
\(=\dfrac{-25x^2-10x-1}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{-\left(5x+1\right)}{x\left(5x-1\right)}\)
c: \(=\dfrac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\dfrac{3y}{x\left(x-3y\right)}\)
\(=\dfrac{x^2+9xy-3xy-9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x^2+6xy-9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
d: \(=\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1+x^2+2x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x^2+8x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{3x^2+9x-x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\)
chứng minh biểu thức M có giá trị không phụ thuộc x,y =)) Giúp mk vs ạ
\(\dfrac{x+9y}{x^2-9y^2}-\dfrac{3y}{x^2+3xy}\)
\(=\dfrac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\dfrac{3y}{x\left(x+3y\right)}\)
\(=\dfrac{x\left(x+9y\right)-3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x-3y}{x\left(x+3y\right)}\)
xin hỏi bạn có viết lộn không, vế trái không có Z mà tại sao vế phải lại xuất hiện Z vậy
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
