Giá trị của biểu thức: A=\(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
Giá trị của biểu thức: A=\(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab+c^2\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab+c^2-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{c^2+ab-c}{\left(a-c\right)\left(b-c\right)}\)
thực hiện phép tính sau\(\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
MTC = (x - y)(x2 + xy + y2)
\(\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
1/x-y-3xy/x^3-y^3+x-y/x^2+xy+y^2
=1/x-y+-3xy/(x-y)(x^2+xy+y^2)+x-y/x^2+xy+y^2
=x^2+xy+y^2/(x-y)(x^2+xy+y^2)+-3xy/(x-y)(x^2+xy+y^2)+x^2-2xy+y^2/(x-y)(x^2+xy+y^2)
=x^2+xy+y^2-3xy+x^2-2xy-y^2/(x-y)(x^2+xy+y^2)
=2x^2-5xy/(x-y)(x^2+xy+y^2)
MTC = (x - y)(x2 + xy + y2)
\(\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
\(\dfrac{-4x^2}{x^2-25}\)-\(\dfrac{2x^2+x}{x^2-25}\)-\(\dfrac{2x}{5-x}\)
Ta có: \(\dfrac{-4x^2}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
\(=\dfrac{-4x^2-2x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x}{x-5}\)
\(=\dfrac{-6x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-6x^2-x+2x^2+10x}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x^2+9x}{\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{-4+25}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
\(\dfrac{-4+25}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
= \(\dfrac{-4+25}{x^2-25}-\dfrac{2x^2+x}{x^2-25}+\dfrac{2x\left(x+5\right)}{x^2-25}\)
= \(\dfrac{-4+25-2x^2-x+2x^2+10x}{x^2-25}\)
= \(\dfrac{21+9x}{x^2-25}\)
x+3/x^2-3x - x/x^2-9
=\(\dfrac{\left(x+3\right)^2}{x\left(x^2-9\right)}-\dfrac{x^2}{x\left(x^2-9\right)}\)
=\(\dfrac{\left(x+3\right)^2-x^2}{x\left(x^2-9\right)}\)
=\(\dfrac{x^2+6x+9-x^2}{x\left(x^2-9\right)}\)
=\(\dfrac{6x+9}{x\left(x^2+9\right)}\)
Tính
\(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}\)
\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
Cho \(A=\dfrac{32x-8x^2+2x^3}{x^3+64}\)(x khác -4)
Tìm x nguyên khi A nguyên
Rút gọn \(A=\dfrac{32x-8x^2+2x^3}{x^3+64}\)
\(A =\frac{32x - 8x^{2} + 2x^{3}}{x^{3}+ 64}\)\(= \frac{2x(16 - 4x + x^{2})}{(x + 4)(x^{2} - 4x + 16)}= \frac{2x(x^{2} - 4x + 16)}{(x + 4)(x^{2} - 4x + 16)}= \frac{2x}{x + 4}\)
\(A=\dfrac{32x-8x^2+2x^3}{x^3+64}\)
\(=\dfrac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
\(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
\(=\dfrac{2x}{x+4}\).
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: \(=\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1+x^2+2x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x^2+8x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{3x^2+9x-x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{3}{x}-\dfrac{1}{x+1}\)
Ta có:
\(\dfrac{3}{x}-\dfrac{1}{x+1}\)
\(=\dfrac{3\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}\)
\(=\dfrac{3x+3-x}{x\left(x+1\right)}\)
\(=\dfrac{2x+3}{x\left(x+1\right)}\)