đúng quy tắc đổi dấu và thực hiện phép tính
\(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
Dùng quy tắc đổi dấu rồi thực hiện các phép tính :
a) \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
b) \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
thực hiện phép tính
a)\(\dfrac{1}{x-5x^2}\)-\(\dfrac{25x-15}{25x^2-1}\)
b)(-\(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}-\dfrac{-1}{4x+16}\))\(\div\dfrac{1}{4x}\)
`a)1/[x-5x^2]-[25x-15]/[25x^2-1]`
`=[-(5x+1)-x(25x-15)]/[x(5x-1)(5x+1)]`
`=[-5x-1-25x^2+15x]/[x(5x-1)(5x+1)]`
`=[-25x^2+10x-1]/[x(5x-1)(5x+1)]`
`=[-(5x-1)^2]/[x(5x-1)(5x+1)]`
`=[1-5x]/[x(5x+1)]`
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`b)(-1/[x^2-4x]+2/[16-x^2]-[-1]/[4x+16]):1/[4x]`
`=[-4(x+4)-8x+x(x-4)]/[4x(x-4)(x+4)].4x`
`=[-4x-16-8x+x^2-4x]/[(x-4)(x+4)]`
`=[x^2-16x-16]/[x^2-16]`
Dùng quy tắc dổi dấu rồi thực hiên các phép tính:
a) \(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}\)
b)\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}\)
a)\(dk,x\ne7;x\ne0\)
\(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}=\frac{\left(4x+13\right)+\left(x-48\right)}{5x\left(x-7\right)}\\ \)
\(=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}\)
b)
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{1-\left(5x\right)^2}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(\frac{1+5x}{x\left(1-5x\right)\left(1+5x\right)}+\frac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-15x+5x+1}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}\)
Dùng qui tắc đổi dấu rồi thực hiện các phép tính: 1 x - 5 x 2 - 25 x - 15 25 x 2 - 1
Thực hiện phép tính:
a, \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(b,\dfrac{2x}{x^2+2xy}-\dfrac{y}{2y^2-xy}+\dfrac{4y}{x^2-4y^2}\)
kha sdaif dòng mik xin phép trình bày bằng lời ạ :
a) tìm MTC rồi quy đồng lên làm bình thường ại , tử cộng tử mấu giữ nguyên
b) cx vậy ạ tách mẫu tìm MTC rồi ....
~ hok tốt ~
Rút gọn biểu thức (chú ý dùng quy tắc đổi dấu để thấy nhân tử chung)
a) \(\dfrac{x+3}{x^2-4}.\dfrac{8-12x+6x^2-x^3}{9x+27}\)
b) \(\dfrac{6x-3}{5x^2+x}.\dfrac{25x^2+10x+1}{1-8x^3}\)
c) \(\dfrac{3x^2-x}{x^2-1}.\dfrac{1-x^4}{\left(1-3x\right)^3}\)
Thực hiện phép tính (chú ý đến quy tắc đổi dấu)
a) \(\dfrac{4\left(x+3\right)}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
b) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
5. a) \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)};\) b) \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
a, \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)
\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)
\(=\dfrac{5x-35}{5x\left(x-7\right)}\)
\(=\dfrac{5\left(x-7\right)}{5x\left(x-7\right)}=\dfrac{1}{x}\)
b, \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x}{x\left(x-5x\right)\left(1+5x\right)}+\dfrac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)}\)\(=\dfrac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}=\dfrac{\left(5x-1\right)^2}{x.\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{\left(5x-1\right)^2}{-x\left(5x-1\right)\left(1+5x\right)}\) \(=\dfrac{-\left(5x-1\right)}{x\left(1+5x\right)}\)
1) thực hiện phép tính
a) \(\dfrac{x-3}{4x+4}-\dfrac{x-1}{6x-30}\)
b) \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
c) \(\dfrac{x+9y}{x^2-9y^2}-\dfrac{3y}{x^2-3xy}\)
d) \(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}-\dfrac{x+3}{1-x^2}\)
e) \(\dfrac{3\left(x-2\right)}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{1-x^2}\)
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: \(=\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1+x^2+2x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x^2+8x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{3x^2+9x-x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\)