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: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
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
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)
a, Rut gon bieu thuc Q
b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)
chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
Chứng minh đẳng thức :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b) \(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3}-x-1\right)\right]:\dfrac{x-1}{x}=\dfrac{2x}{x-1}\)
c) \(\left[\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
giải phương trình
a, \(\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1}\)
b,\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
c,\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
d, \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
e, \(x^3+x^2+x+1=0\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
