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a, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)
\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)
b, Ta có : \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)
TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)
c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)
\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )
e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )
TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)
a) \(B=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+5\right)\left(x-1\right)}\right)\)\(:\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{9-3x+x^2+10x+25-\left(x^2-1\right)}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7x+35}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7\left(x+5\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)\cdot7\left(x-2\right)}=\frac{x+1}{x-2}\)
b) \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\)
\(\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow x^2-4x+5x-20=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\left(KTM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Với x = 4 ta có \(B=\frac{4+1}{4-2}=\frac{5}{2}\)
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3
