thuc hien phep tinh:
a,2xyz+4xyz-\(\dfrac{1}{2}\)xyz
b,\(\dfrac{x^2}{2}\)+\(\dfrac{x^2}{3}\)+\(\dfrac{x^2}{4}\)
thuc hien phep tinh:
a,2xyz+4xyz-\(\dfrac{1}{2}\)xyz
b,\(\dfrac{x^2}{2}\)+\(\dfrac{x^2}{3}\)+\(\dfrac{x^2}{4}\)
2xyz+4xyz-\(\frac{1}{2}\) xyz
=(2+4-\(\frac{1}{2}\) )(xxxyyyzzz)
=3.5x3 y3z3
bai 1 thuc hien phep tinh a)\(\dfrac{\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}\right)}{\left(\dfrac{1}{x+2}+\dfrac{1}{x-2}\right)}\)
b)\(\left(\dfrac{5x+y}{x^2-5xy}+\dfrac{5x-y}{x^2+5xy}\right)\cdot\dfrac{x^2-25y^2}{x^2+y^2}\)
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
thuc hien phep tinh:
a) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\)
\(=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
thuc hien phep tinh:
a) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x}\)
Sửa đề: \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\)
\(=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
1) Thuc hien phep tinh cong 2 phan thuc \(\dfrac{2x}{x^2-2x+1}+\dfrac{x+1}{x-1}\) duoc ket qua la:
A. \(\dfrac{x^2+2x+1}{\left(x-1\right)^2}\) B. \(\dfrac{x^2+2x-1}{\left(x-1\right)^2}\) C. \(\dfrac{x^2-x-1}{\left(x-1\right)^2}\) D. \(\dfrac{x^2-2x-1}{\left(x-1\right)^2}\)
thuc hien phep tinh
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
Thuc hien phep tinh :\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)
\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)
=\(\dfrac{18x}{x\left(x^2-9\right)}-\dfrac{\left(2-x\right)\left(x-3\right)x}{\left(x+3\right)\left(x-3\right)x}-\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)x}\)
=\(\dfrac{18x-\left(2-x\right)\left(x-3\right)x-3x\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-\left(2x-6-x^2+3x\right)x-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{18x-2x^2+6x+x^3-3x^2-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)
=\(\dfrac{x^3-8x^2+15x}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x^2-8x+15\right)}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2-3x-5x+15}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x\left(x-3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x-5\right)}{x+3}\)
thuc hien phep tinh
\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(a.\)
\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
\(=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
\(b.\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\dfrac{5\left(x-1\right)}{3\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x-1\right)}\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}=\dfrac{x}{3\left(x-1\right)}\)\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
thuc hien phep tinh
a.\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b.\(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+1-2\right)\)
c.\(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d.\(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)