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\(\frac{4.\left(x+3\right)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4.\left(x+3\right)}{\left(3x-1\right)}\cdot\frac{\left(3x-1\right)}{x^2+3x}=\frac{4.\left(x+3\right)}{x.\left(x+3\right)}=\frac{4}{x}\)
\(a,\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)
\(=\frac{-x-2}{1-x}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)
\(=\frac{-x-2}{1-x}+\frac{-\left(x-9\right)}{1-x}+\frac{-\left(x-9\right)}{1-x}\)
\(=\frac{-x-2-x+9-x+9}{1-x}=\frac{-3x+16}{1-x}\)
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) \(\frac{4(x+3)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4(x+3)}{3x-1}.\frac{3x-1}{x^2+3x}=\frac{4(x+3)}{x^2+3x}=\frac{4(x+3)}{x(x+3)}=\frac{4}{x}\)
chứng minh biểu thức M có giá trị không phụ thuộc x,y =)) Giúp mk vs ạ
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: \(=\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1+x^2+2x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x^2+8x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{3x^2+9x-x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\)
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0
xin hỏi bạn có viết lộn không, vế trái không có Z mà tại sao vế phải lại xuất hiện Z vậy
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona