Question

Cho 2 biểu thức A=x+3/x+2 và B=x/x+1+2/x-1-4/x^2-1

1)Tính giá trị của biểu thức A khi x=-3

2)Chứng minh B=x+2/x+1

3)Đặt M=A.B.tÌM x để m>0

Answer 1

a, Do \(x=-3\)\(=>A=\frac{x+3}{x+2}=\frac{-3+3}{-3+2}=\frac{0}{-1}=0\)

Vậy A = 0 khi x = -3

b, Ta có : \(B=\frac{x}{x+1}+\frac{2}{x-1}-\frac{4}{x^2-1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{4}{x^2-1}\)

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

\(=\frac{x+2}{x+1}\)(đpcm)

Answer 2

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Answer 3

Nó tố cáo mày giờ 

Answer 4

Ta có \(B=\frac{x+2}{x+1}\); \(A=\frac{x+3}{x+2}\)

\(=>M=A.B=\frac{x+3}{x+2}.\frac{x+2}{x+1}=\frac{x+3}{x+1}\)

Để \(M>0=>\frac{x+3}{x+1}>0=>\orbr{\begin{cases}x+3>0;x+1>0\\x+3< 0;x+1< 0\end{cases}}\)

\(< =>\orbr{\begin{cases}x>-1\\x< -3\end{cases}}\)

Vậy để M > 0 thì x > -1 hoặc x < -3