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a) \(\frac{2x-1}{x^2-5x+6}\)
\(=\frac{5x-10-3x+9}{x^2-2x-3x+6}\)
\(=\frac{5\left(x-2\right)-3\left(x-3\right)}{x\left(x-2\right)-3\left(x-2\right)}\)
\(=\frac{5\left(x-2\right)-3\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{5\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)\(-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{5}{x-3}+\frac{-3}{x-2}\)
\(M=\frac{2x-1}{x^2-5x+6}=\frac{2x-1}{\left(x-2\right)\left(x-3\right)}=\frac{5\left(x-2\right)-3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{x-3}-\frac{3}{x-2}=\frac{5}{x-3}+\frac{3}{2-x}\)
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
\(5x^2+10xy=5x\left(x+2y\right)\)
\(x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)
\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)
a) = \(\frac{2x}{\left(x-2\right)\left(x-3\right)}\)-\(\frac{1}{\left(x-2\right)\left(x-3\right)}\)
các bài sau tt
k hiểu