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a: \(M=\dfrac{-y+4}{y-2}+\dfrac{1}{y-2}+\dfrac{3}{y+2}\)
\(=\dfrac{-y+5}{y-2}+\dfrac{3}{y+2}=\dfrac{-y^2-2y+5y+10+3y-6}{\left(y-2\right)\left(y+2\right)}\)
\(=\dfrac{-y^2+6y+4}{\left(y-2\right)\left(y+2\right)}\)
b: Khi y=3 thì \(M=\dfrac{-3^2+6\cdot3+4}{\left(3-2\right)\left(3+2\right)}=\dfrac{-5+18}{5}=\dfrac{13}{5}\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
Bài 2:
a: \(\Leftrightarrow4x^2-4x+1-4x^2-16x-16=9\)
=>-20x-15=9
=>-20x=24
=>x=-6/5
b: \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
=>9x=18
=>x=2
a, \(M=\frac{xy^2+y^2\left(y^2-x\right)+1}{x^2y^4+2y^4+x^2+2}=\frac{y^2\left(x+y^2-x\right)+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}=\frac{y^4+1}{\left(y^4+1\right)\left(x^2+2\right)}=\frac{1}{x^2+2}\)
Thay x=-3 vào M
=>\(M=\frac{1}{\left(-3\right)^2+2}=\frac{1}{11}\)
b, Vì \(x^2\ge0\Rightarrow x^2+2\ge2\Rightarrow M=\frac{1}{x^2+2}>0\)
để M xác định
\(\Rightarrow\orbr{\begin{cases}y-1\ne0\\y+1\ne0\end{cases}}\Rightarrow\frac{y\ne1}{y\ne-1}.\)
\(b,M=\frac{1}{y-1}+\frac{y}{y+1}+\frac{2y^2}{y^2-1}\)
\(M=\frac{y+1}{\left(y+1\right)\left(y-1\right)}+\frac{y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)}+\frac{2y^2}{\left(y+1\right)\left(y-1\right)}\)
\(M=\frac{y+1-y^2+y+2y^2}{\left(y+1\right)\left(y-1\right)}=\frac{1+2y+y^2}{\left(y+1\right)\left(y-1\right)}=\frac{\left(1+y\right)^2}{\left(y+1\right)\left(y-1\right)}\)
\(M=\frac{y+1}{y-1}\)
c, Để M nhận giá trị nguyên
\(\Rightarrow y+1⋮y-1\)
\(\Leftrightarrow y-1+2⋮y-1\)
\(\Rightarrow y-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
y = .... Tự tính
