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a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
a) \(\left(3-2x\right)\left(x+1\right)+x\left(2x-1\right)=3x+3-2x^2-2x+2x^2-x=3\)
b) \(\frac{x^2+9}{x^2+3x}+\frac{6}{x+3}=\frac{x^2+9}{x\left(x+3\right)}+\frac{6x}{x\left(x+3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)}=\frac{x+3}{x}\)
c)\(\frac{2+x}{2-x}+\frac{4x^2}{4-x^2}+\frac{x-2}{2+x}=\frac{\left(x+2\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}+\frac{-\left(x-2\right)^2}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}=\frac{4x^2+8x}{\left(x+2\right)\left(2-x\right)}=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}=\frac{4x}{2-x}\)
d) \(\left(x^3+4x^2+6x+4\right):\left(x+2\right)\)
\(=\left(x^3+2x^2+2x^2+4x+2x+4\right):\left(x+2\right)\)
\(=\left[x^2\left(x+2\right)+2x\left(x+2\right)+2\left(x+2\right)\right]:\left(x+2\right)\)
\(=\left(x^2+2x+2\right)\left(x+2\right):\left(x+2\right)=x^2+2x+2\)
a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{x+1}{x}\div\frac{x^2-1}{x}=\frac{x+1}{x}\cdot\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x-1}\)
b) \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right)\div\left(\frac{1}{x+2}+\frac{1}{x-2}\right)=\frac{\left(x-2\right)^2-\left(x+2^2\right)}{\left(x^2-4\right)^2}\div\frac{x-2+x+2}{x^2-4}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x^2-4\right)^2}\cdot\frac{x^2-4}{2x}=\frac{2x\cdot\left(-4\right)}{x^2-4}\cdot\frac{1}{2x}=\frac{-4}{x^2-4}\)
a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x^2-1}{x}}=\frac{x+1}{x}\cdot\frac{x}{x^2-1}=\frac{1}{x-1}\)
b) \(\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2^2\right)}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(\Leftrightarrow\left(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(\Leftrightarrow\left(\frac{x^2-4x+4-x^2-4x-4}{\left[\left(x-2\right)\left(x+2\right)\right]^2}\right):\left(\frac{x-2+x+2}{x^2-4}\right)\)
\(\Leftrightarrow\frac{-8x}{\left(x^2-4\right)^2}\cdot\frac{x^2-4}{2x}\)\(\Leftrightarrow-\frac{4}{x^2-4}\)
d) \(\frac{3x}{x^3-1}+\frac{x-1}{x^2+x+1}\Leftrightarrow\frac{3x}{x^3-1}+\frac{\left(x-1\right)^2}{x^3-1}\)
\(\Leftrightarrow\frac{x^2-2x+1+3x}{x^3-1}=\frac{x^2+x+1}{x^3-1}=\frac{1}{x-1}\)
còn lại chút giải tiếp !!!