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\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(=\frac{\left(x+y\right)^2-1}{\left(x-1\right)^2-y^2}\)
\(=\frac{\left(x+y-1\right)\left(x+y+1\right)}{\left(x-1-y\right)\left(x-1+y\right)}\)
\(=\frac{x+y+1}{x-y-1}\)
MTC : ( x - 1 )( x2 + x + 1 )
Ta có : \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
Hnay mới học thì hnay trả lời nhá :P
\(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1}\)
Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2+x+1=x^2+x+1\)
MTC : \(\left(x-1\right)\left(x^2+x+1\right)\)
\(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
câu c nè
\(\frac{x^2-3x+5}{x+1}=\frac{\left(x^2+2x+1\right)-5x+4}{x+1}=\frac{\left(x+1\right)^2-5\left(x+1\right)+9}{x+1}\)
Ta có \(\frac{x+2}{x+1}=\frac{\left(x+1\right)+1}{x+1}=1+\frac{1}{x+1}\)
Vì đa thức \(f\left(x\right)=2x^2+\text{ax}+b-2.\)chia hết cho \(x-1;x+2\)
Theo đinh lí Bơ - ru ta có
\(f\left(1\right)=2.1^2+a+b-2=0\Rightarrow a+b=0\) (1)
\(f\left(-2\right)=2.\left(-2\right)^2-2a+b-2=0\Rightarrow6-2a+b=0\Rightarrow2a-b=6\)(2)
Cộng (1) và (2) suy ra
\(3a=6\Rightarrow a=2\)thay vào a+b=0 ta có : \(2+b=0\Rightarrow b=-2\)
Vậy a=2 ; a=-2
Bài 2:
a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)
\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)
\(\frac{x+2}{x+1}=\frac{x}{x+1}+\frac{2}{x+1}\)
\(\frac{2x-3}{x-1}=\frac{2x}{x-1}+\frac{-3}{x-1}\)
\(\frac{x^2-3x+5}{x+1}=\frac{x^2}{x+1}+\frac{-3x+5}{x+1}\)
