Câu hỏi của Ling ling 2k7

Question

Thu gọn Q hoặc P giúp mk zới

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Nguyễn Lê Phước Thịnh · Accepted Answer

1:$Q=\frac{3-x}{x-2}+1=\frac{3-x+x-2}{x-2}=\frac{1}{x-2}$

a: |x|=2

=>x=2(loại) hoặc x=-2(nhận)

Thay x=-2 vào Q, ta được:

$Q=\frac{3-\left(-2\right)}{-2-2}+1=\frac{3+2}{-4}+1=\frac{5}{-4}+1=\frac{-5}{4}+1=\frac{-5+4}{4}=-\frac14$

b: $x=\sqrt{7-4\sqrt3}$

=>$x=\sqrt{\left(2-\sqrt3\right)^2}=2-\sqrt3$

Thay $x=2-\sqrt3$ vào Q, ta được:

$Q=\frac{1}{2-\sqrt3-2}=\frac{1}{-\sqrt3}=\frac{-\sqrt3}{3}$

2: Q<0

=>$\frac{1}{x-2}<0$

=>x-2<0

=>x<2

Q>1

=>Q-1>0

=>$\frac{1}{x-2}-1>0$

=>$\frac{1-x+2}{x-2}>0$

=>$\frac{3-x}{x-2}>0$

=>$\frac{x-3}{x-2}<0$

=>2

3: $P=\frac{x}{x-2}-\frac{x+1}{x+2}-\frac{x+4}{x^2-4}$

$=\frac{x\left(x+2\right)-\left(x+1\right)\left(x-2\right)-x-4}{\left(x-2\right)\left(x+2\right)}$

$=\frac{x^2+2x-\left(x^2-x-2\right)-x-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+x-4-x^2+x+2}{\left(x-2\right)\left(x+2\right)}$

$=\frac{2x-2}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}$

S=P:Q

$=\frac{2\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}:\frac{1}{x-2}=\frac{2\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\cdot\left(x-2\right)=\frac{2\left(x-1\right)}{x+2}$

4: $S=\frac{2\left(x-1\right)}{x+2}=\frac{2x-2}{x+2}=\frac{2x+4-6}{x+2}=2-\frac{6}{x+2}$

Để S nguyên thì -6⋮x+2

=>x+2∈{1;-1;2;-2;3;-3;6;-6}

=>x∈{-1;-3;0;-4;1;-5;4;-8}

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