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a: \(Q=-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1\)
\(A=x^2y-3x+1-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{2}x^2y-\dfrac{7}{12}xy^2-3x\)
b: \(P=\dfrac{3}{4}xy^2+\dfrac{4}{9}x-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{6}xy^2+\dfrac{16}{9}x-\dfrac{1}{2}x^2y-1\)
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a: ĐKXĐ: x<>2; x<>3
\(Q=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Để P<1 thì P-1<0
=>\(\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
=>x<3
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a: Để M có nghĩa thì x+2<>0
hay x<>-2
b: Để M=4 thì 4x+8=3x+9
hay x=1(nhận)
c: Để M là số nguyên thì \(3x+9⋮x+2\)
\(\Leftrightarrow3x+6+3⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{-1;-3;1;-5\right\}\)
`a)`
Để `M` có nghĩa thì:
`x+2\ne0`
`<=>x\ne-2`
Vậy `x\ne-2` thì `M` có nghĩa
`b)`
`M=4`
`<=>(3x+9)/(x+2)=4`
`=>4x+8=3x+9`
`<=>4x-3x=9-8`
`<=>x=1`
Vậy `x=1` thì `M=4`
`c)`
`M\inZZ<=>(3x+9)/(x+2)\inZZ`
`=>3x+9\vdotsx+2`
`=>3x+6+3\vdotsx+2`
`=>3.(x+2)+3\vdotsx+2`
`=>x+2\in Ư(3)={+-1;+-3}`
Ta có bảng:
$\begin{array}{|c|c|}\hline x+2&1&-1&3&-3\\\hline x&-1&-3&1&-5\\\hline\end{array}$
Vậy `x\in{-1;-3;1;-5}` thì `M\inZZ`
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Ta có : \(A=\frac{x+9}{x+3}=\frac{\left(x+3\right)+6}{x+3}=1+\frac{6}{x+3}\)
Để \(A\in Z\)thì \(\frac{6}{x+3}\in Z\)
\(\Rightarrow6⋮x+3\Rightarrow x+3\inƯ_{\left(6\right)}=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Lập bảng ra ok :)))