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 ...
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
ĐKXĐ : \(x\ne\pm3\)
a) \(P=\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\)
\(P=\left(\frac{x^2+1}{\left(x+3\right)\left(x-3\right)}-\frac{x}{x+3}-\frac{5}{x-3}\right):\left(\frac{2x+10}{x+3}-\frac{x+3}{x+3}\right)\)
\(P=\left(\frac{x^2+1}{\left(x+3\right)\left(x-3\right)}-\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x+10-x-3}{x+3}\right)\)
\(P=\left(\frac{x^2+1-x\left(x-3\right)-5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+7}{x+3}\right)\)
\(P=\left(\frac{x^2+1-x^2+3x-5x-15}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+7}{x+3}\right)\)
\(P=\frac{-2x-14}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+7}\)
\(P=\frac{-2\left(x+7\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+7}\)
\(P=-\frac{2}{x-3}\)
b) \(P=-\frac{2}{x-3}\)( ĐKXĐ : \(x\ne3\))
x - 1 = 2 => x = 3 ( không tmđk )
Vậy không có giá trị của P khi x - 1 = 2
c) \(P=\frac{x+5}{6}\Leftrightarrow\frac{-2}{x-3}=\frac{x+5}{6}\)
<=> -2.6 = ( x - 3 )( x + 5 )
<=> -12 = x2 + 2x - 15
<=> x2 + 2x - 15 + 12 = 0
<=> x2 + 2x - 3 = 0
<=> x2 - x + 3x - 3 = 0
<=> x( x - 1 ) + 3( x - 1 ) = 0
<=> ( x + 3 )( x - 1 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
Vậy x = { -3 ; 1 }
mình bổ sụng thêm điều kiện \(x\ne-7\)để \(\frac{2x+10}{x+3}-1\ne0\)
