Các bạn chỉ cần giúp mik câu c ạ

\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

\(=\dfrac{x^3-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)

\(=\dfrac{x^3-x^2-2x+2x-4}{x^2-4}\)

\(=\dfrac{x^3-x^2-4}{x^2-4}\)

a,\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=x-1\)

b, C=0\(\Rightarrow x-1=0\Rightarrow x=1\)

c, Để C nhận giá trị dương thì \(x-1\ge0\Rightarrow x\ge1\)

1.football

2.hockey

3.table tennis

4.golf

5.volleyball

6.squash

7.basketball

8.cricket

9.tennis

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xin loi minh ko biet

xin loi minh ko biet

ko bik

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ko bik

Câu 45: B

Câu 45:B

B

$M_O = 16(đvC)$

$PTK = Y + 16.2 = M_{Cu} = 64(đvC) \Rightarrow Y = 32$
Vậy Y là lưu huỳnh, KHHH : S

d) Ta có: \(1,25\cdot\left(-2\right)^2-0.5:\dfrac{5}{12}+15\%\)

\(=\dfrac{5}{4}\cdot4-\dfrac{1}{2}\cdot\dfrac{12}{5}+\dfrac{3}{20}\)

\(=5-\dfrac{6}{5}+\dfrac{3}{20}\)

\(=\dfrac{100}{20}-\dfrac{120}{20}+\dfrac{3}{20}\)

\(=\dfrac{-17}{20}\)

More progress in gender equality will be made by the Vietnamese government

Old men should be treated with respect by children

2 câu dưới mình làm rồi nha bạn =))