Question

[công chúa xinh đẹp]

CHO BIỂU THỨC P=\(\text{[}\frac{X^2+1}{X^2-9}-\frac{X}{X+3}+\frac{5}{3-X}\text{]}\div\text{[}\frac{2X+10}{X+3}-1\text{]}\)1]

A ,RÚT GỌN P

B.TÍNH P KHI [X-1]=2

C.TÌM X ĐỂ P=\(\frac{X+5}{6}\)

Answer 1

[Edogawa Conan]

ĐKXĐ: x \(\ne\)\(\pm\)3; x \(\ne\)-7

a) Ta có: P = \(\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\)

P = \(\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2x+10-x-3}{x+3}\right)\)

P = \(\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}:\frac{x+7}{x+3}\)

P = \(\frac{-2x-14}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)

P = \(\frac{-2\left(x+7\right)}{x-3}\cdot\frac{1}{x+7}=-\frac{2}{x-3}\)

b) Với x \(\ne\)\(\pm\)3 và x \(\ne\)-7

Ta có: x - 1 = 2 <=> x = 3 (ktm)

=> ko tồn tại giá trị P khi x - 1 = 2

c) Với x \(\ne\)\(\pm\)3; và x \(\ne\)-7

Ta có: P = \(\frac{x+5}{6}\)

<=> \(-\frac{2}{x-3}=\frac{x+5}{6}\)

=> (x - 3)(x + 5) = -12

<=> x² + 2x - 15 = -12

<=> x² + 2x - 3 = 0

<=> x²  + 3x - x - 3 = 0

<=> (x - 1)(x + 3) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)

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

Vậy ...

Answer 2

[Tran Le Khanh Linh]

a) \(P=\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\left(x\ne\pm3\right)\)

\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x}{x+3}-\frac{5}{x-3}\right):\frac{2x+10-x-3}{x+3}\)

\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x+15}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+7}{x+3}\)

\(=\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)

\(=\frac{\left(-2x-14\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}\)

\(=\frac{-2\left(x+7\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}=-\frac{2}{x-3}\)

vậy \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)

b) ta có \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)

có x-1=2 

<=> x=3 (không thỏa mãn điều kiện)

vậy không có giá trị P để x-1=2

c) ta có: \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)

P=\(\frac{x+5}{6}\)=> \(\frac{-2}{x-3}=\frac{x+5}{6}\)

\(\Leftrightarrow x^2+2x-15=-12\)

\(\Leftrightarrow x^2+2x-3=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)

đối chiếu điều kiện ta thấy x=1 thỏa mãn điều kiện

vậy \(P=\frac{x+5}{6}\)đạt được khi x=1